The Normal Distribution

The Normal Distribution

The normal distribution, also known as the Gaussian distribution, is the most widely-used general purpose distribution. It is for this reason that it is included among the lifetime distributions commonly used for reliability and life data analysis. There are some who argue that the normal distribution is inappropriate for modeling lifetime data because the left-hand limit of the distribution extends to negative infinity. This could conceivably result in modeling negative times-to-failure. However, provided that the distribution in question has a relatively high mean and a relatively small standard deviation, the issue of negative failure times should not present itself as a problem. Nevertheless, the normal distribution has been shown to be useful for modeling the lifetimes of consumable items, such as printer toner cartridges.

Normal Probability Density Function

The [math]\displaystyle{ pdf }[/math] of the normal distribution is given by:

[math]\displaystyle{ f(t)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{t-\mu }{\sigma } \right)}^{2}}}} }[/math]

where:

[math]\displaystyle{ \mu }[/math] = mean of the normal times-to-faiure, also noted as [math]\displaystyle{ \bar{T} }[/math],

[math]\displaystyle{ \theta=\text{standard deviation of the times-to-failure} }[/math]

It is a two-parameter distribution with parameters [math]\displaystyle{ \mu }[/math] (or [math]\displaystyle{ \bar{T} }[/math] ) and [math]\displaystyle{ {{\sigma }} }[/math] , i.e. the mean and the standard deviation, respectively.

Normal Statistical Properties

The Normal Mean, Median and Mode

The normal mean or MTTF is actually one of the parameters of the distribution, usually denoted as [math]\displaystyle{ \mu . }[/math] Since the normal distribution is symmetrical, the median and the mode are always equal to the mean,

[math]\displaystyle{ \mu =\tilde{T}=\breve{T} }[/math]

The Normal Standard Deviation

As with the mean, the standard deviation for the normal distribution is actually one of the parameters, usually denoted as [math]\displaystyle{ {{\sigma }_{T}}. }[/math]

The Normal Reliability Function

The reliability for a mission of time [math]\displaystyle{ T }[/math] for the normal distribution is determined by:

[math]\displaystyle{ R(t)=\int_{t}^{\infty }f(x)dx=\int_{t}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx }[/math]

There is no closed-form solution for the normal reliability function. Solutions can be obtained via the use of standard normal tables. Since the application automatically solves for the reliability, we will not discuss manual solution methods. For interested readers, full explanations can be found in the references.

The Normal Conditional Reliability Function

The normal conditional reliability function is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{\int_{T+t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}{\int_{T}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Once again, the use of standard normal tables for the calculation of the normal conditional reliability is necessary, as there is no closed form solution.

The Normal Reliable Life

Since there is no closed-form solution for the normal reliability function, there will also be no closed-form solution for the normal reliable life. To determine the normal reliable life, one must solve:

[math]\displaystyle{ R(T)=\int_{T}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}dt }[/math]

for [math]\displaystyle{ T }[/math] .

The Normal Failure Rate Function

The instantaneous normal failure rate is given by:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}}{\int_{t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Characteristics of the Normal Distribution

Some of the specific characteristics of the normal distribution are the following:

• The normal [math]\displaystyle{ pdf }[/math] has a mean, [math]\displaystyle{ \bar{T} }[/math] , which is equal to the median, [math]\displaystyle{ \breve{T} }[/math], and also equal to the mode, [math]\displaystyle{ \tilde{T} }[/math], or [math]\displaystyle{ \bar{T}=\breve{T}=\tilde{T} }[/math]. This is because the normal distribution is symmetrical about its mean.

• The mean, [math]\displaystyle{ \mu }[/math] , or the mean life or the [math]\displaystyle{ MTTF }[/math] , is also the location parameter of the normal [math]\displaystyle{ pdf }[/math] , as it locates the [math]\displaystyle{ pdf }[/math] along the abscissa. It can assume values of [math]\displaystyle{ -\infty \lt \bar{T}\lt \infty }[/math] .
• The normal [math]\displaystyle{ pdf }[/math] has no shape parameter. This means that the normal [math]\displaystyle{ pdf }[/math] has only one shape, the bell shape, and this shape does not change.

• The standard deviation, [math]\displaystyle{ {{\sigma }} }[/math] , is the scale parameter of the normal [math]\displaystyle{ pdf }[/math] .

• As [math]\displaystyle{ {{\sigma }} }[/math] decreases, the [math]\displaystyle{ pdf }[/math] gets pushed toward the mean, or it becomes narrower and taller.

• As [math]\displaystyle{ {{\sigma }} }[/math] increases, the [math]\displaystyle{ pdf }[/math] spreads out away from the mean, or it becomes broader and shallower.

• The standard deviation can assume values of [math]\displaystyle{ 0\lt {{\sigma }}\lt \infty }[/math] .

• The greater the variability, the larger the value of [math]\displaystyle{ {{\sigma }} }[/math] , and vice versa.

• The standard deviation is also the distance between the mean and the point of inflection of the [math]\displaystyle{ pdf }[/math] , on each side of the mean. The point of inflection is that point of the [math]\displaystyle{ pdf }[/math] where the slope changes its value from a decreasing to an increasing one, or where the second derivative of the [math]\displaystyle{ pdf }[/math] has a value of zero.

• The normal [math]\displaystyle{ pdf }[/math] starts at [math]\displaystyle{ t=-\infty }[/math] with an [math]\displaystyle{ f(t)=0 }[/math] . As [math]\displaystyle{ t }[/math] increases, [math]\displaystyle{ f(t) }[/math] also increases, goes through its point of inflection and reaches its maximum value at [math]\displaystyle{ t=\bar{T} }[/math] . Thereafter, [math]\displaystyle{ f(t) }[/math] decreases, goes through its point of inflection, and assumes a value of [math]\displaystyle{ f(t)=0 }[/math] at [math]\displaystyle{ t=+\infty }[/math] .

Weibull++ Notes on Negative Time Values

One of the disadvantages of using the normal distribution for reliability calculations is the fact that the normal distribution starts at negative infinity. This can result in negative values for some of the results. Negative values for time are not accepted in most of the components of Weibull++, nor are they implemented. Certain components of the application reserve negative values for suspensions, or will not return negative results. For example, the Quick Calculation Pad will return a null value (zero) if the result is negative. Only the Free-Form (Probit) data sheet can accept negative values for the random variable (x-axis values).

Estimation of the Parameters

Probability Plotting

As described before, probability plotting involves plotting the failure times and associated unreliability estimates on specially constructed probability plotting paper. The form of this paper is based on a linearization of the [math]\displaystyle{ cdf }[/math] of the specific distribution. For the normal distribution, the cumulative density function can be written as:

[math]\displaystyle{ F(t)=\Phi \left( \frac{t-\mu }{{{\sigma }}} \right) }[/math]

or:

[math]\displaystyle{ {{\Phi }^{-1}}\left[ F(t) \right]=-\frac{\mu}{\sigma}+\frac{1}{\sigma}t }[/math]

where:

[math]\displaystyle{ \Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt }[/math]

Now, let:

[math]\displaystyle{ y={{\Phi }^{-1}}\left[ F(t) \right] }[/math]

[math]\displaystyle{ a=-\frac{\mu }{\sigma } }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\sigma } }[/math]

which results in the linear equation of:

[math]\displaystyle{ y=a+bT }[/math]

The normal probability paper resulting from this linearized [math]\displaystyle{ cdf }[/math] function is shown next.

Since the normal distribution is symmetrical, the area under the [math]\displaystyle{ pdf }[/math] curve from [math]\displaystyle{ -\infty }[/math] to [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ 0.5 }[/math] , as is the area from [math]\displaystyle{ \mu }[/math] to [math]\displaystyle{ +\infty }[/math] . Consequently, the value of [math]\displaystyle{ \mu }[/math] is said to be the point where [math]\displaystyle{ R(t)=Q(t)=50% }[/math] . This means that the estimate of [math]\displaystyle{ \mu }[/math] can be read from the point where the plotted line crosses the 50% unreliability line.

To determine the value of [math]\displaystyle{ \sigma }[/math] from the probability plot, it is first necessary to understand that the area under the [math]\displaystyle{ pdf }[/math] curve that lies between one standard deviation in either direction from the mean (or two standard deviations total) represents 68.3% of the area under the curve. This is represented graphically in the following figure.

Consequently, the interval between [math]\displaystyle{ Q(t)=84.15% }[/math] and [math]\displaystyle{ Q(t)=15.85% }[/math] represents two standard deviations, since this is an interval of 68.3% ( [math]\displaystyle{ 84.15-15.85=68.3 }[/math] ), and is centered on the mean at 50%. As a result, the standard deviation can be estimated from:

[math]\displaystyle{ \widehat{\sigma }=\frac{t(Q=84.15%)-t(Q=15.85%)}{2} }[/math]

That is: the value of [math]\displaystyle{ \widehat{\sigma } }[/math] is obtained by subtracting the time value where the plotted line crosses the 84.15% unreliability line from the time value where the plotted line crosses the 15.85% unreliability line and dividing the result by two. This process is illustrated in the following example.

Example 1:

The Normal Distribution

The normal distribution, also known as the Gaussian distribution, is the most widely-used general purpose distribution. It is for this reason that it is included among the lifetime distributions commonly used for reliability and life data analysis. There are some who argue that the normal distribution is inappropriate for modeling lifetime data because the left-hand limit of the distribution extends to negative infinity. This could conceivably result in modeling negative times-to-failure. However, provided that the distribution in question has a relatively high mean and a relatively small standard deviation, the issue of negative failure times should not present itself as a problem. Nevertheless, the normal distribution has been shown to be useful for modeling the lifetimes of consumable items, such as printer toner cartridges.

Normal Probability Density Function

The [math]\displaystyle{ pdf }[/math] of the normal distribution is given by:

[math]\displaystyle{ f(t)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{t-\mu }{\sigma } \right)}^{2}}}} }[/math]

where:

[math]\displaystyle{ \mu }[/math] = mean of the normal times-to-faiure, also noted as [math]\displaystyle{ \bar{T} }[/math],

[math]\displaystyle{ \theta=\text{standard deviation of the times-to-failure} }[/math]

It is a two-parameter distribution with parameters [math]\displaystyle{ \mu }[/math] (or [math]\displaystyle{ \bar{T} }[/math] ) and [math]\displaystyle{ {{\sigma }} }[/math] , i.e. the mean and the standard deviation, respectively.

Normal Statistical Properties

The Normal Mean, Median and Mode

The normal mean or MTTF is actually one of the parameters of the distribution, usually denoted as [math]\displaystyle{ \mu . }[/math] Since the normal distribution is symmetrical, the median and the mode are always equal to the mean,

[math]\displaystyle{ \mu =\tilde{T}=\breve{T} }[/math]

The Normal Standard Deviation

As with the mean, the standard deviation for the normal distribution is actually one of the parameters, usually denoted as [math]\displaystyle{ {{\sigma }_{T}}. }[/math]

The Normal Reliability Function

The reliability for a mission of time [math]\displaystyle{ T }[/math] for the normal distribution is determined by:

[math]\displaystyle{ R(t)=\int_{t}^{\infty }f(x)dx=\int_{t}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx }[/math]

There is no closed-form solution for the normal reliability function. Solutions can be obtained via the use of standard normal tables. Since the application automatically solves for the reliability, we will not discuss manual solution methods. For interested readers, full explanations can be found in the references.

The Normal Conditional Reliability Function

The normal conditional reliability function is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{\int_{T+t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}{\int_{T}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Once again, the use of standard normal tables for the calculation of the normal conditional reliability is necessary, as there is no closed form solution.

The Normal Reliable Life

Since there is no closed-form solution for the normal reliability function, there will also be no closed-form solution for the normal reliable life. To determine the normal reliable life, one must solve:

[math]\displaystyle{ R(T)=\int_{T}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}dt }[/math]

for [math]\displaystyle{ T }[/math] .

The Normal Failure Rate Function

The instantaneous normal failure rate is given by:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}}{\int_{t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Characteristics of the Normal Distribution

Some of the specific characteristics of the normal distribution are the following:

• The normal [math]\displaystyle{ pdf }[/math] has a mean, [math]\displaystyle{ \bar{T} }[/math] , which is equal to the median, [math]\displaystyle{ \breve{T} }[/math], and also equal to the mode, [math]\displaystyle{ \tilde{T} }[/math], or [math]\displaystyle{ \bar{T}=\breve{T}=\tilde{T} }[/math]. This is because the normal distribution is symmetrical about its mean.

• The mean, [math]\displaystyle{ \mu }[/math] , or the mean life or the [math]\displaystyle{ MTTF }[/math] , is also the location parameter of the normal [math]\displaystyle{ pdf }[/math] , as it locates the [math]\displaystyle{ pdf }[/math] along the abscissa. It can assume values of [math]\displaystyle{ -\infty \lt \bar{T}\lt \infty }[/math] .
• The normal [math]\displaystyle{ pdf }[/math] has no shape parameter. This means that the normal [math]\displaystyle{ pdf }[/math] has only one shape, the bell shape, and this shape does not change.

• The standard deviation, [math]\displaystyle{ {{\sigma }} }[/math] , is the scale parameter of the normal [math]\displaystyle{ pdf }[/math] .

• As [math]\displaystyle{ {{\sigma }} }[/math] decreases, the [math]\displaystyle{ pdf }[/math] gets pushed toward the mean, or it becomes narrower and taller.

• As [math]\displaystyle{ {{\sigma }} }[/math] increases, the [math]\displaystyle{ pdf }[/math] spreads out away from the mean, or it becomes broader and shallower.

• The standard deviation can assume values of [math]\displaystyle{ 0\lt {{\sigma }}\lt \infty }[/math] .

• The greater the variability, the larger the value of [math]\displaystyle{ {{\sigma }} }[/math] , and vice versa.

• The standard deviation is also the distance between the mean and the point of inflection of the [math]\displaystyle{ pdf }[/math] , on each side of the mean. The point of inflection is that point of the [math]\displaystyle{ pdf }[/math] where the slope changes its value from a decreasing to an increasing one, or where the second derivative of the [math]\displaystyle{ pdf }[/math] has a value of zero.

• The normal [math]\displaystyle{ pdf }[/math] starts at [math]\displaystyle{ t=-\infty }[/math] with an [math]\displaystyle{ f(t)=0 }[/math] . As [math]\displaystyle{ t }[/math] increases, [math]\displaystyle{ f(t) }[/math] also increases, goes through its point of inflection and reaches its maximum value at [math]\displaystyle{ t=\bar{T} }[/math] . Thereafter, [math]\displaystyle{ f(t) }[/math] decreases, goes through its point of inflection, and assumes a value of [math]\displaystyle{ f(t)=0 }[/math] at [math]\displaystyle{ t=+\infty }[/math] .

Weibull++ Notes on Negative Time Values

One of the disadvantages of using the normal distribution for reliability calculations is the fact that the normal distribution starts at negative infinity. This can result in negative values for some of the results. Negative values for time are not accepted in most of the components of Weibull++, nor are they implemented. Certain components of the application reserve negative values for suspensions, or will not return negative results. For example, the Quick Calculation Pad will return a null value (zero) if the result is negative. Only the Free-Form (Probit) data sheet can accept negative values for the random variable (x-axis values).

Estimation of the Parameters

Probability Plotting

As described before, probability plotting involves plotting the failure times and associated unreliability estimates on specially constructed probability plotting paper. The form of this paper is based on a linearization of the [math]\displaystyle{ cdf }[/math] of the specific distribution. For the normal distribution, the cumulative density function can be written as:

[math]\displaystyle{ F(t)=\Phi \left( \frac{t-\mu }{{{\sigma }}} \right) }[/math]

or:

[math]\displaystyle{ {{\Phi }^{-1}}\left[ F(t) \right]=-\frac{\mu}{\sigma}+\frac{1}{\sigma}t }[/math]

where:

[math]\displaystyle{ \Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt }[/math]

Now, let:

[math]\displaystyle{ y={{\Phi }^{-1}}\left[ F(t) \right] }[/math]

[math]\displaystyle{ a=-\frac{\mu }{\sigma } }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\sigma } }[/math]

which results in the linear equation of:

[math]\displaystyle{ y=a+bT }[/math]

The normal probability paper resulting from this linearized [math]\displaystyle{ cdf }[/math] function is shown next.

Since the normal distribution is symmetrical, the area under the [math]\displaystyle{ pdf }[/math] curve from [math]\displaystyle{ -\infty }[/math] to [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ 0.5 }[/math] , as is the area from [math]\displaystyle{ \mu }[/math] to [math]\displaystyle{ +\infty }[/math] . Consequently, the value of [math]\displaystyle{ \mu }[/math] is said to be the point where [math]\displaystyle{ R(t)=Q(t)=50% }[/math] . This means that the estimate of [math]\displaystyle{ \mu }[/math] can be read from the point where the plotted line crosses the 50% unreliability line.

To determine the value of [math]\displaystyle{ \sigma }[/math] from the probability plot, it is first necessary to understand that the area under the [math]\displaystyle{ pdf }[/math] curve that lies between one standard deviation in either direction from the mean (or two standard deviations total) represents 68.3% of the area under the curve. This is represented graphically in the following figure.

Consequently, the interval between [math]\displaystyle{ Q(t)=84.15% }[/math] and [math]\displaystyle{ Q(t)=15.85% }[/math] represents two standard deviations, since this is an interval of 68.3% ( [math]\displaystyle{ 84.15-15.85=68.3 }[/math] ), and is centered on the mean at 50%. As a result, the standard deviation can be estimated from:

[math]\displaystyle{ \widehat{\sigma }=\frac{t(Q=84.15%)-t(Q=15.85%)}{2} }[/math]

That is: the value of [math]\displaystyle{ \widehat{\sigma } }[/math] is obtained by subtracting the time value where the plotted line crosses the 84.15% unreliability line from the time value where the plotted line crosses the 15.85% unreliability line and dividing the result by two. This process is illustrated in the following example.

Example 1:

Rank Regression on Y

Performing rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

The least squares parameter estimation method (regression analysis) was discussed in Parameter Estimation, and the following equations for regression on Y were derived:

[math]\displaystyle{ \begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\ =& \frac{\sum_{i=1}^N y_{i}}{N}-\hat{b}\frac{\sum_{i=1}^{N}x_{i}}{N}\\ \end{align} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}} }[/math]

In the case of the normal distribution, the equations for [math]\displaystyle{ {{y}_{i}} }[/math] and [math]\displaystyle{ {{x}_{i}} }[/math] are:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

where the values for [math]\displaystyle{ F({{T}_{i}}) }[/math] are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, [math]\displaystyle{ \widehat{\sigma } }[/math] and [math]\displaystyle{ \widehat{\mu } }[/math] can easily be obtained from above equations.

The Correlation Coefficient The estimator of the sample correlation coefficient, [math]\displaystyle{ \hat{\rho } }[/math] , is given by:

[math]\displaystyle{ \hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}} }[/math]

Example 2:

Rank Regression on X

As was mentioned previously, performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the fitted line is minimized.

Again, the first task is to bring our function, the probability of failure function for normal distribution, into a linear form. This step is exactly the same as in regression on Y analysis. All other equations apply in this case as they did for the regression on Y. The deviation from the previous analysis begins on the least squares fit step where: in this case, we treat [math]\displaystyle{ x }[/math] as the dependent variable and [math]\displaystyle{ y }[/math] as the independent variable. The best-fitting straight line for the data, for regression on X, is the straight line:

[math]\displaystyle{ x=\widehat{a}+\widehat{b}y }[/math]

The corresponding equations for [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are:

[math]\displaystyle{ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{N}} }[/math]

where:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

and the [math]\displaystyle{ F({{t}_{i}}) }[/math] values are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, solve the above linear equation for the unknown value of [math]\displaystyle{ y }[/math] which corresponds to:

[math]\displaystyle{ y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x }[/math]

Solving for the parameters, we get:

[math]\displaystyle{ a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{\mu }{\sigma }\Rightarrow \mu =\widehat{a} }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\widehat{b}}=\frac{1}{\sigma }\Rightarrow \sigma =\widehat{b} }[/math]

The correlation coefficient is evaluated as before.

Example 3:

The following examples illustrate the different types of life data that can be analyzed in Weibull++ using the normal distribution. For more information on the different types of life data, see Life Data Classification.

Complete Data Example

6 units are tested to failure. The following failure times data are obtained: 12125, 11260, 12080, 12825, 13550 and 14670 hours. Assuming that the data are normally distributed, do the following:

Objectives

1. Find the parameters for the data set, using the Rank Regression on X (RRX) parameter estimation method

2. Obtain the probability plot for the data with 90%, two-sided Type 1 confidence bounds.

3. Obtain the pdf plot for the data.

4. Using the Quick Calculation Pad (QCP), determine the reliability for a mission of 11,000 hours, as well as the upper and lower two-sided 90% confidence limit on this reliability.

5. Using the QCP, determine the MTTF, as well as the upper and lower two-sided 90% confidence limit on this MTTF.

6. Obtain tabulated values for the failure rate for 10 different mission end times. The mission end times are 1,000 to 10,000 hours, using increments of 1,000 hours.

Solution

The following figure shows the data as entered in Weibull++, as well as the calculated parameters.

The following figures show the probability plot with the 90% two-sided confidence bounds and the pdf plot.

Both the reliability and MTTF can be easily obtained from the QCP. The QCP, with results, for both cases is shown in the next two figures.

To obtain tabulated values for the failure rate, use the Analysis Workbook or General Spreadsheet features that are included in Weibull++. (For more information on these features, please refer to the Weibull++ User's Guide. For a step-by-step example on creating Weibull++ reports, please see the Quick Start Guide). The following worksheet shows the mission times and the corresponding failure rates.

Suspension Data Example

19 units are being reliability tested and the following is a table of their times-to-failure and suspensions.

Non-Grouped Data Times-to-Failure Data with Suspensions
Data point index	Last Inspected	State End Time
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.07 \\ & {{{\hat{\sigma }}}_{T}}= & 28.41. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 46.40 \\ & {{{\hat{\sigma }}}_{T}}= & 28.64. \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 47.34 \\ & {{{\hat{\sigma }}}_{T}}= & 29.96. \end{align}\,\! }[/math]

Interval Censored Data Example

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Interval Data
Data point index	Last Inspected	State End Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

This is a sequence of interval times-to-failure data. Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 7.740. \end{align}\,\! }[/math]

For rank regression on x:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 9.03. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on y (RRY) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.39 \\ & {{{\hat{\sigma }}}_{T}}= & 9.25. \end{align}\,\! }[/math]

The following plot shows the results if the data were analyzed using the rank regression on X (RRX) method.

Mixed Data Types Example

Suppose our data set includes left and right censored, interval censored and complete data, as shown in the following table.

Grouped Data Times-to-Failure with Suspensions and Intervals (Interval, Left and Right Censored)
Data point index	Number in State	Last Inspection	State (S or F)	State End Time
1	1	10	F	10
2	1	20	S	20
3	2	0	F	30
4	2	40	F	40
5	1	50	F	50
6	1	60	S	60
7	1	70	F	70
8	2	20	F	80
9	1	10	F	85
10	1	100	F	100

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.11 \\ & {{{\hat{\sigma }}}_{T}}= & 26.42 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 49.99 \\ & {{{\hat{\sigma }}}_{T}}= & 30.17 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 51.61 \\ & {{{\hat{\sigma }}}_{T}}= & 33.07 \end{align}\,\! }[/math]

Comparison of Analysis Methods

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Times-to-Failure Data
Data point index	State F or S	State End Time
1	F	2
2	F	5
3	F	11
4	F	23
5	F	29
6	F	37
7	F	43
8	F	59

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 18.57 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 21.64 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 22.28. \end{align}\,\! }[/math]

Rank Regression on Y

Performing rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

The least squares parameter estimation method (regression analysis) was discussed in Parameter Estimation, and the following equations for regression on Y were derived:

[math]\displaystyle{ \begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\ =& \frac{\sum_{i=1}^N y_{i}}{N}-\hat{b}\frac{\sum_{i=1}^{N}x_{i}}{N}\\ \end{align} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}} }[/math]

In the case of the normal distribution, the equations for [math]\displaystyle{ {{y}_{i}} }[/math] and [math]\displaystyle{ {{x}_{i}} }[/math] are:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

where the values for [math]\displaystyle{ F({{T}_{i}}) }[/math] are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, [math]\displaystyle{ \widehat{\sigma } }[/math] and [math]\displaystyle{ \widehat{\mu } }[/math] can easily be obtained from above equations.

The Correlation Coefficient The estimator of the sample correlation coefficient, [math]\displaystyle{ \hat{\rho } }[/math] , is given by:

[math]\displaystyle{ \hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}} }[/math]

Example 2:

The Normal Distribution

The normal distribution, also known as the Gaussian distribution, is the most widely-used general purpose distribution. It is for this reason that it is included among the lifetime distributions commonly used for reliability and life data analysis. There are some who argue that the normal distribution is inappropriate for modeling lifetime data because the left-hand limit of the distribution extends to negative infinity. This could conceivably result in modeling negative times-to-failure. However, provided that the distribution in question has a relatively high mean and a relatively small standard deviation, the issue of negative failure times should not present itself as a problem. Nevertheless, the normal distribution has been shown to be useful for modeling the lifetimes of consumable items, such as printer toner cartridges.

Normal Probability Density Function

The [math]\displaystyle{ pdf }[/math] of the normal distribution is given by:

[math]\displaystyle{ f(t)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{t-\mu }{\sigma } \right)}^{2}}}} }[/math]

where:

[math]\displaystyle{ \mu }[/math] = mean of the normal times-to-faiure, also noted as [math]\displaystyle{ \bar{T} }[/math],

[math]\displaystyle{ \theta=\text{standard deviation of the times-to-failure} }[/math]

It is a two-parameter distribution with parameters [math]\displaystyle{ \mu }[/math] (or [math]\displaystyle{ \bar{T} }[/math] ) and [math]\displaystyle{ {{\sigma }} }[/math] , i.e. the mean and the standard deviation, respectively.

Normal Statistical Properties

The Normal Mean, Median and Mode

The normal mean or MTTF is actually one of the parameters of the distribution, usually denoted as [math]\displaystyle{ \mu . }[/math] Since the normal distribution is symmetrical, the median and the mode are always equal to the mean,

[math]\displaystyle{ \mu =\tilde{T}=\breve{T} }[/math]

The Normal Standard Deviation

As with the mean, the standard deviation for the normal distribution is actually one of the parameters, usually denoted as [math]\displaystyle{ {{\sigma }_{T}}. }[/math]

The Normal Reliability Function

The reliability for a mission of time [math]\displaystyle{ T }[/math] for the normal distribution is determined by:

[math]\displaystyle{ R(t)=\int_{t}^{\infty }f(x)dx=\int_{t}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx }[/math]

There is no closed-form solution for the normal reliability function. Solutions can be obtained via the use of standard normal tables. Since the application automatically solves for the reliability, we will not discuss manual solution methods. For interested readers, full explanations can be found in the references.

The Normal Conditional Reliability Function

The normal conditional reliability function is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{\int_{T+t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}{\int_{T}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Once again, the use of standard normal tables for the calculation of the normal conditional reliability is necessary, as there is no closed form solution.

The Normal Reliable Life

Since there is no closed-form solution for the normal reliability function, there will also be no closed-form solution for the normal reliable life. To determine the normal reliable life, one must solve:

[math]\displaystyle{ R(T)=\int_{T}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}dt }[/math]

for [math]\displaystyle{ T }[/math] .

The Normal Failure Rate Function

The instantaneous normal failure rate is given by:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}}{\int_{t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Characteristics of the Normal Distribution

Some of the specific characteristics of the normal distribution are the following:

• The normal [math]\displaystyle{ pdf }[/math] has a mean, [math]\displaystyle{ \bar{T} }[/math] , which is equal to the median, [math]\displaystyle{ \breve{T} }[/math], and also equal to the mode, [math]\displaystyle{ \tilde{T} }[/math], or [math]\displaystyle{ \bar{T}=\breve{T}=\tilde{T} }[/math]. This is because the normal distribution is symmetrical about its mean.

• The mean, [math]\displaystyle{ \mu }[/math] , or the mean life or the [math]\displaystyle{ MTTF }[/math] , is also the location parameter of the normal [math]\displaystyle{ pdf }[/math] , as it locates the [math]\displaystyle{ pdf }[/math] along the abscissa. It can assume values of [math]\displaystyle{ -\infty \lt \bar{T}\lt \infty }[/math] .
• The normal [math]\displaystyle{ pdf }[/math] has no shape parameter. This means that the normal [math]\displaystyle{ pdf }[/math] has only one shape, the bell shape, and this shape does not change.

• The standard deviation, [math]\displaystyle{ {{\sigma }} }[/math] , is the scale parameter of the normal [math]\displaystyle{ pdf }[/math] .

• As [math]\displaystyle{ {{\sigma }} }[/math] decreases, the [math]\displaystyle{ pdf }[/math] gets pushed toward the mean, or it becomes narrower and taller.

• As [math]\displaystyle{ {{\sigma }} }[/math] increases, the [math]\displaystyle{ pdf }[/math] spreads out away from the mean, or it becomes broader and shallower.

• The standard deviation can assume values of [math]\displaystyle{ 0\lt {{\sigma }}\lt \infty }[/math] .

• The greater the variability, the larger the value of [math]\displaystyle{ {{\sigma }} }[/math] , and vice versa.

• The standard deviation is also the distance between the mean and the point of inflection of the [math]\displaystyle{ pdf }[/math] , on each side of the mean. The point of inflection is that point of the [math]\displaystyle{ pdf }[/math] where the slope changes its value from a decreasing to an increasing one, or where the second derivative of the [math]\displaystyle{ pdf }[/math] has a value of zero.

• The normal [math]\displaystyle{ pdf }[/math] starts at [math]\displaystyle{ t=-\infty }[/math] with an [math]\displaystyle{ f(t)=0 }[/math] . As [math]\displaystyle{ t }[/math] increases, [math]\displaystyle{ f(t) }[/math] also increases, goes through its point of inflection and reaches its maximum value at [math]\displaystyle{ t=\bar{T} }[/math] . Thereafter, [math]\displaystyle{ f(t) }[/math] decreases, goes through its point of inflection, and assumes a value of [math]\displaystyle{ f(t)=0 }[/math] at [math]\displaystyle{ t=+\infty }[/math] .

Weibull++ Notes on Negative Time Values

One of the disadvantages of using the normal distribution for reliability calculations is the fact that the normal distribution starts at negative infinity. This can result in negative values for some of the results. Negative values for time are not accepted in most of the components of Weibull++, nor are they implemented. Certain components of the application reserve negative values for suspensions, or will not return negative results. For example, the Quick Calculation Pad will return a null value (zero) if the result is negative. Only the Free-Form (Probit) data sheet can accept negative values for the random variable (x-axis values).

Estimation of the Parameters

Probability Plotting

As described before, probability plotting involves plotting the failure times and associated unreliability estimates on specially constructed probability plotting paper. The form of this paper is based on a linearization of the [math]\displaystyle{ cdf }[/math] of the specific distribution. For the normal distribution, the cumulative density function can be written as:

[math]\displaystyle{ F(t)=\Phi \left( \frac{t-\mu }{{{\sigma }}} \right) }[/math]

or:

[math]\displaystyle{ {{\Phi }^{-1}}\left[ F(t) \right]=-\frac{\mu}{\sigma}+\frac{1}{\sigma}t }[/math]

where:

[math]\displaystyle{ \Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt }[/math]

Now, let:

[math]\displaystyle{ y={{\Phi }^{-1}}\left[ F(t) \right] }[/math]

[math]\displaystyle{ a=-\frac{\mu }{\sigma } }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\sigma } }[/math]

which results in the linear equation of:

[math]\displaystyle{ y=a+bT }[/math]

The normal probability paper resulting from this linearized [math]\displaystyle{ cdf }[/math] function is shown next.

Since the normal distribution is symmetrical, the area under the [math]\displaystyle{ pdf }[/math] curve from [math]\displaystyle{ -\infty }[/math] to [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ 0.5 }[/math] , as is the area from [math]\displaystyle{ \mu }[/math] to [math]\displaystyle{ +\infty }[/math] . Consequently, the value of [math]\displaystyle{ \mu }[/math] is said to be the point where [math]\displaystyle{ R(t)=Q(t)=50% }[/math] . This means that the estimate of [math]\displaystyle{ \mu }[/math] can be read from the point where the plotted line crosses the 50% unreliability line.

To determine the value of [math]\displaystyle{ \sigma }[/math] from the probability plot, it is first necessary to understand that the area under the [math]\displaystyle{ pdf }[/math] curve that lies between one standard deviation in either direction from the mean (or two standard deviations total) represents 68.3% of the area under the curve. This is represented graphically in the following figure.

Consequently, the interval between [math]\displaystyle{ Q(t)=84.15% }[/math] and [math]\displaystyle{ Q(t)=15.85% }[/math] represents two standard deviations, since this is an interval of 68.3% ( [math]\displaystyle{ 84.15-15.85=68.3 }[/math] ), and is centered on the mean at 50%. As a result, the standard deviation can be estimated from:

[math]\displaystyle{ \widehat{\sigma }=\frac{t(Q=84.15%)-t(Q=15.85%)}{2} }[/math]

That is: the value of [math]\displaystyle{ \widehat{\sigma } }[/math] is obtained by subtracting the time value where the plotted line crosses the 84.15% unreliability line from the time value where the plotted line crosses the 15.85% unreliability line and dividing the result by two. This process is illustrated in the following example.

Example 1:

Rank Regression on Y

Performing rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

The least squares parameter estimation method (regression analysis) was discussed in Parameter Estimation, and the following equations for regression on Y were derived:

[math]\displaystyle{ \begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\ =& \frac{\sum_{i=1}^N y_{i}}{N}-\hat{b}\frac{\sum_{i=1}^{N}x_{i}}{N}\\ \end{align} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}} }[/math]

In the case of the normal distribution, the equations for [math]\displaystyle{ {{y}_{i}} }[/math] and [math]\displaystyle{ {{x}_{i}} }[/math] are:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

where the values for [math]\displaystyle{ F({{T}_{i}}) }[/math] are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, [math]\displaystyle{ \widehat{\sigma } }[/math] and [math]\displaystyle{ \widehat{\mu } }[/math] can easily be obtained from above equations.

The Correlation Coefficient The estimator of the sample correlation coefficient, [math]\displaystyle{ \hat{\rho } }[/math] , is given by:

[math]\displaystyle{ \hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}} }[/math]

Example 2:

Rank Regression on X

As was mentioned previously, performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the fitted line is minimized.

Again, the first task is to bring our function, the probability of failure function for normal distribution, into a linear form. This step is exactly the same as in regression on Y analysis. All other equations apply in this case as they did for the regression on Y. The deviation from the previous analysis begins on the least squares fit step where: in this case, we treat [math]\displaystyle{ x }[/math] as the dependent variable and [math]\displaystyle{ y }[/math] as the independent variable. The best-fitting straight line for the data, for regression on X, is the straight line:

[math]\displaystyle{ x=\widehat{a}+\widehat{b}y }[/math]

The corresponding equations for [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are:

[math]\displaystyle{ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{N}} }[/math]

where:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

and the [math]\displaystyle{ F({{t}_{i}}) }[/math] values are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, solve the above linear equation for the unknown value of [math]\displaystyle{ y }[/math] which corresponds to:

[math]\displaystyle{ y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x }[/math]

Solving for the parameters, we get:

[math]\displaystyle{ a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{\mu }{\sigma }\Rightarrow \mu =\widehat{a} }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\widehat{b}}=\frac{1}{\sigma }\Rightarrow \sigma =\widehat{b} }[/math]

The correlation coefficient is evaluated as before.

Example 3:

The following examples illustrate the different types of life data that can be analyzed in Weibull++ using the normal distribution. For more information on the different types of life data, see Life Data Classification.

Complete Data Example

6 units are tested to failure. The following failure times data are obtained: 12125, 11260, 12080, 12825, 13550 and 14670 hours. Assuming that the data are normally distributed, do the following:

Objectives

1. Find the parameters for the data set, using the Rank Regression on X (RRX) parameter estimation method

2. Obtain the probability plot for the data with 90%, two-sided Type 1 confidence bounds.

3. Obtain the pdf plot for the data.

4. Using the Quick Calculation Pad (QCP), determine the reliability for a mission of 11,000 hours, as well as the upper and lower two-sided 90% confidence limit on this reliability.

5. Using the QCP, determine the MTTF, as well as the upper and lower two-sided 90% confidence limit on this MTTF.

6. Obtain tabulated values for the failure rate for 10 different mission end times. The mission end times are 1,000 to 10,000 hours, using increments of 1,000 hours.

Solution

The following figure shows the data as entered in Weibull++, as well as the calculated parameters.

The following figures show the probability plot with the 90% two-sided confidence bounds and the pdf plot.

Both the reliability and MTTF can be easily obtained from the QCP. The QCP, with results, for both cases is shown in the next two figures.

To obtain tabulated values for the failure rate, use the Analysis Workbook or General Spreadsheet features that are included in Weibull++. (For more information on these features, please refer to the Weibull++ User's Guide. For a step-by-step example on creating Weibull++ reports, please see the Quick Start Guide). The following worksheet shows the mission times and the corresponding failure rates.

Suspension Data Example

19 units are being reliability tested and the following is a table of their times-to-failure and suspensions.

Non-Grouped Data Times-to-Failure Data with Suspensions
Data point index	Last Inspected	State End Time
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.07 \\ & {{{\hat{\sigma }}}_{T}}= & 28.41. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 46.40 \\ & {{{\hat{\sigma }}}_{T}}= & 28.64. \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 47.34 \\ & {{{\hat{\sigma }}}_{T}}= & 29.96. \end{align}\,\! }[/math]

Interval Censored Data Example

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Interval Data
Data point index	Last Inspected	State End Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

This is a sequence of interval times-to-failure data. Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 7.740. \end{align}\,\! }[/math]

For rank regression on x:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 9.03. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on y (RRY) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.39 \\ & {{{\hat{\sigma }}}_{T}}= & 9.25. \end{align}\,\! }[/math]

The following plot shows the results if the data were analyzed using the rank regression on X (RRX) method.

Mixed Data Types Example

Suppose our data set includes left and right censored, interval censored and complete data, as shown in the following table.

Grouped Data Times-to-Failure with Suspensions and Intervals (Interval, Left and Right Censored)
Data point index	Number in State	Last Inspection	State (S or F)	State End Time
1	1	10	F	10
2	1	20	S	20
3	2	0	F	30
4	2	40	F	40
5	1	50	F	50
6	1	60	S	60
7	1	70	F	70
8	2	20	F	80
9	1	10	F	85
10	1	100	F	100

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.11 \\ & {{{\hat{\sigma }}}_{T}}= & 26.42 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 49.99 \\ & {{{\hat{\sigma }}}_{T}}= & 30.17 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 51.61 \\ & {{{\hat{\sigma }}}_{T}}= & 33.07 \end{align}\,\! }[/math]

Comparison of Analysis Methods

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Times-to-Failure Data
Data point index	State F or S	State End Time
1	F	2
2	F	5
3	F	11
4	F	23
5	F	29
6	F	37
7	F	43
8	F	59

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 18.57 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 21.64 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 22.28. \end{align}\,\! }[/math]

Rank Regression on X

As was mentioned previously, performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the fitted line is minimized.

Again, the first task is to bring our function, the probability of failure function for normal distribution, into a linear form. This step is exactly the same as in regression on Y analysis. All other equations apply in this case as they did for the regression on Y. The deviation from the previous analysis begins on the least squares fit step where: in this case, we treat [math]\displaystyle{ x }[/math] as the dependent variable and [math]\displaystyle{ y }[/math] as the independent variable. The best-fitting straight line for the data, for regression on X, is the straight line:

[math]\displaystyle{ x=\widehat{a}+\widehat{b}y }[/math]

The corresponding equations for [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are:

[math]\displaystyle{ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{N}} }[/math]

where:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

and the [math]\displaystyle{ F({{t}_{i}}) }[/math] values are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, solve the above linear equation for the unknown value of [math]\displaystyle{ y }[/math] which corresponds to:

[math]\displaystyle{ y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x }[/math]

Solving for the parameters, we get:

[math]\displaystyle{ a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{\mu }{\sigma }\Rightarrow \mu =\widehat{a} }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\widehat{b}}=\frac{1}{\sigma }\Rightarrow \sigma =\widehat{b} }[/math]

The correlation coefficient is evaluated as before.

Example 3:

The Normal Distribution

The normal distribution, also known as the Gaussian distribution, is the most widely-used general purpose distribution. It is for this reason that it is included among the lifetime distributions commonly used for reliability and life data analysis. There are some who argue that the normal distribution is inappropriate for modeling lifetime data because the left-hand limit of the distribution extends to negative infinity. This could conceivably result in modeling negative times-to-failure. However, provided that the distribution in question has a relatively high mean and a relatively small standard deviation, the issue of negative failure times should not present itself as a problem. Nevertheless, the normal distribution has been shown to be useful for modeling the lifetimes of consumable items, such as printer toner cartridges.

Normal Probability Density Function

The [math]\displaystyle{ pdf }[/math] of the normal distribution is given by:

[math]\displaystyle{ f(t)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{t-\mu }{\sigma } \right)}^{2}}}} }[/math]

where:

[math]\displaystyle{ \mu }[/math] = mean of the normal times-to-faiure, also noted as [math]\displaystyle{ \bar{T} }[/math],

[math]\displaystyle{ \theta=\text{standard deviation of the times-to-failure} }[/math]

It is a two-parameter distribution with parameters [math]\displaystyle{ \mu }[/math] (or [math]\displaystyle{ \bar{T} }[/math] ) and [math]\displaystyle{ {{\sigma }} }[/math] , i.e. the mean and the standard deviation, respectively.

Normal Statistical Properties

The Normal Mean, Median and Mode

The normal mean or MTTF is actually one of the parameters of the distribution, usually denoted as [math]\displaystyle{ \mu . }[/math] Since the normal distribution is symmetrical, the median and the mode are always equal to the mean,

[math]\displaystyle{ \mu =\tilde{T}=\breve{T} }[/math]

The Normal Standard Deviation

As with the mean, the standard deviation for the normal distribution is actually one of the parameters, usually denoted as [math]\displaystyle{ {{\sigma }_{T}}. }[/math]

The Normal Reliability Function

The reliability for a mission of time [math]\displaystyle{ T }[/math] for the normal distribution is determined by:

[math]\displaystyle{ R(t)=\int_{t}^{\infty }f(x)dx=\int_{t}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx }[/math]

There is no closed-form solution for the normal reliability function. Solutions can be obtained via the use of standard normal tables. Since the application automatically solves for the reliability, we will not discuss manual solution methods. For interested readers, full explanations can be found in the references.

The Normal Conditional Reliability Function

The normal conditional reliability function is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{\int_{T+t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}{\int_{T}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Once again, the use of standard normal tables for the calculation of the normal conditional reliability is necessary, as there is no closed form solution.

The Normal Reliable Life

Since there is no closed-form solution for the normal reliability function, there will also be no closed-form solution for the normal reliable life. To determine the normal reliable life, one must solve:

[math]\displaystyle{ R(T)=\int_{T}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}dt }[/math]

for [math]\displaystyle{ T }[/math] .

The Normal Failure Rate Function

The instantaneous normal failure rate is given by:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}}{\int_{t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Characteristics of the Normal Distribution

Some of the specific characteristics of the normal distribution are the following:

• The normal [math]\displaystyle{ pdf }[/math] has a mean, [math]\displaystyle{ \bar{T} }[/math] , which is equal to the median, [math]\displaystyle{ \breve{T} }[/math], and also equal to the mode, [math]\displaystyle{ \tilde{T} }[/math], or [math]\displaystyle{ \bar{T}=\breve{T}=\tilde{T} }[/math]. This is because the normal distribution is symmetrical about its mean.

• The mean, [math]\displaystyle{ \mu }[/math] , or the mean life or the [math]\displaystyle{ MTTF }[/math] , is also the location parameter of the normal [math]\displaystyle{ pdf }[/math] , as it locates the [math]\displaystyle{ pdf }[/math] along the abscissa. It can assume values of [math]\displaystyle{ -\infty \lt \bar{T}\lt \infty }[/math] .
• The normal [math]\displaystyle{ pdf }[/math] has no shape parameter. This means that the normal [math]\displaystyle{ pdf }[/math] has only one shape, the bell shape, and this shape does not change.

• The standard deviation, [math]\displaystyle{ {{\sigma }} }[/math] , is the scale parameter of the normal [math]\displaystyle{ pdf }[/math] .

• As [math]\displaystyle{ {{\sigma }} }[/math] decreases, the [math]\displaystyle{ pdf }[/math] gets pushed toward the mean, or it becomes narrower and taller.

• As [math]\displaystyle{ {{\sigma }} }[/math] increases, the [math]\displaystyle{ pdf }[/math] spreads out away from the mean, or it becomes broader and shallower.

• The standard deviation can assume values of [math]\displaystyle{ 0\lt {{\sigma }}\lt \infty }[/math] .

• The greater the variability, the larger the value of [math]\displaystyle{ {{\sigma }} }[/math] , and vice versa.

• The standard deviation is also the distance between the mean and the point of inflection of the [math]\displaystyle{ pdf }[/math] , on each side of the mean. The point of inflection is that point of the [math]\displaystyle{ pdf }[/math] where the slope changes its value from a decreasing to an increasing one, or where the second derivative of the [math]\displaystyle{ pdf }[/math] has a value of zero.

• The normal [math]\displaystyle{ pdf }[/math] starts at [math]\displaystyle{ t=-\infty }[/math] with an [math]\displaystyle{ f(t)=0 }[/math] . As [math]\displaystyle{ t }[/math] increases, [math]\displaystyle{ f(t) }[/math] also increases, goes through its point of inflection and reaches its maximum value at [math]\displaystyle{ t=\bar{T} }[/math] . Thereafter, [math]\displaystyle{ f(t) }[/math] decreases, goes through its point of inflection, and assumes a value of [math]\displaystyle{ f(t)=0 }[/math] at [math]\displaystyle{ t=+\infty }[/math] .

Weibull++ Notes on Negative Time Values

One of the disadvantages of using the normal distribution for reliability calculations is the fact that the normal distribution starts at negative infinity. This can result in negative values for some of the results. Negative values for time are not accepted in most of the components of Weibull++, nor are they implemented. Certain components of the application reserve negative values for suspensions, or will not return negative results. For example, the Quick Calculation Pad will return a null value (zero) if the result is negative. Only the Free-Form (Probit) data sheet can accept negative values for the random variable (x-axis values).

Estimation of the Parameters

Probability Plotting

As described before, probability plotting involves plotting the failure times and associated unreliability estimates on specially constructed probability plotting paper. The form of this paper is based on a linearization of the [math]\displaystyle{ cdf }[/math] of the specific distribution. For the normal distribution, the cumulative density function can be written as:

[math]\displaystyle{ F(t)=\Phi \left( \frac{t-\mu }{{{\sigma }}} \right) }[/math]

or:

[math]\displaystyle{ {{\Phi }^{-1}}\left[ F(t) \right]=-\frac{\mu}{\sigma}+\frac{1}{\sigma}t }[/math]

where:

[math]\displaystyle{ \Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt }[/math]

Now, let:

[math]\displaystyle{ y={{\Phi }^{-1}}\left[ F(t) \right] }[/math]

[math]\displaystyle{ a=-\frac{\mu }{\sigma } }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\sigma } }[/math]

which results in the linear equation of:

[math]\displaystyle{ y=a+bT }[/math]

The normal probability paper resulting from this linearized [math]\displaystyle{ cdf }[/math] function is shown next.

Since the normal distribution is symmetrical, the area under the [math]\displaystyle{ pdf }[/math] curve from [math]\displaystyle{ -\infty }[/math] to [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ 0.5 }[/math] , as is the area from [math]\displaystyle{ \mu }[/math] to [math]\displaystyle{ +\infty }[/math] . Consequently, the value of [math]\displaystyle{ \mu }[/math] is said to be the point where [math]\displaystyle{ R(t)=Q(t)=50% }[/math] . This means that the estimate of [math]\displaystyle{ \mu }[/math] can be read from the point where the plotted line crosses the 50% unreliability line.

To determine the value of [math]\displaystyle{ \sigma }[/math] from the probability plot, it is first necessary to understand that the area under the [math]\displaystyle{ pdf }[/math] curve that lies between one standard deviation in either direction from the mean (or two standard deviations total) represents 68.3% of the area under the curve. This is represented graphically in the following figure.

Consequently, the interval between [math]\displaystyle{ Q(t)=84.15% }[/math] and [math]\displaystyle{ Q(t)=15.85% }[/math] represents two standard deviations, since this is an interval of 68.3% ( [math]\displaystyle{ 84.15-15.85=68.3 }[/math] ), and is centered on the mean at 50%. As a result, the standard deviation can be estimated from:

[math]\displaystyle{ \widehat{\sigma }=\frac{t(Q=84.15%)-t(Q=15.85%)}{2} }[/math]

That is: the value of [math]\displaystyle{ \widehat{\sigma } }[/math] is obtained by subtracting the time value where the plotted line crosses the 84.15% unreliability line from the time value where the plotted line crosses the 15.85% unreliability line and dividing the result by two. This process is illustrated in the following example.

Example 1:

Rank Regression on Y

Performing rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

The least squares parameter estimation method (regression analysis) was discussed in Parameter Estimation, and the following equations for regression on Y were derived:

[math]\displaystyle{ \begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\ =& \frac{\sum_{i=1}^N y_{i}}{N}-\hat{b}\frac{\sum_{i=1}^{N}x_{i}}{N}\\ \end{align} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}} }[/math]

In the case of the normal distribution, the equations for [math]\displaystyle{ {{y}_{i}} }[/math] and [math]\displaystyle{ {{x}_{i}} }[/math] are:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

where the values for [math]\displaystyle{ F({{T}_{i}}) }[/math] are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, [math]\displaystyle{ \widehat{\sigma } }[/math] and [math]\displaystyle{ \widehat{\mu } }[/math] can easily be obtained from above equations.

The Correlation Coefficient The estimator of the sample correlation coefficient, [math]\displaystyle{ \hat{\rho } }[/math] , is given by:

[math]\displaystyle{ \hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}} }[/math]

Example 2:

Rank Regression on X

As was mentioned previously, performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the fitted line is minimized.

Again, the first task is to bring our function, the probability of failure function for normal distribution, into a linear form. This step is exactly the same as in regression on Y analysis. All other equations apply in this case as they did for the regression on Y. The deviation from the previous analysis begins on the least squares fit step where: in this case, we treat [math]\displaystyle{ x }[/math] as the dependent variable and [math]\displaystyle{ y }[/math] as the independent variable. The best-fitting straight line for the data, for regression on X, is the straight line:

[math]\displaystyle{ x=\widehat{a}+\widehat{b}y }[/math]

The corresponding equations for [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are:

[math]\displaystyle{ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{N}} }[/math]

where:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

and the [math]\displaystyle{ F({{t}_{i}}) }[/math] values are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, solve the above linear equation for the unknown value of [math]\displaystyle{ y }[/math] which corresponds to:

[math]\displaystyle{ y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x }[/math]

Solving for the parameters, we get:

[math]\displaystyle{ a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{\mu }{\sigma }\Rightarrow \mu =\widehat{a} }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\widehat{b}}=\frac{1}{\sigma }\Rightarrow \sigma =\widehat{b} }[/math]

The correlation coefficient is evaluated as before.

Example 3:

The following examples illustrate the different types of life data that can be analyzed in Weibull++ using the normal distribution. For more information on the different types of life data, see Life Data Classification.

Complete Data Example

6 units are tested to failure. The following failure times data are obtained: 12125, 11260, 12080, 12825, 13550 and 14670 hours. Assuming that the data are normally distributed, do the following:

Objectives

1. Find the parameters for the data set, using the Rank Regression on X (RRX) parameter estimation method

2. Obtain the probability plot for the data with 90%, two-sided Type 1 confidence bounds.

3. Obtain the pdf plot for the data.

4. Using the Quick Calculation Pad (QCP), determine the reliability for a mission of 11,000 hours, as well as the upper and lower two-sided 90% confidence limit on this reliability.

5. Using the QCP, determine the MTTF, as well as the upper and lower two-sided 90% confidence limit on this MTTF.

6. Obtain tabulated values for the failure rate for 10 different mission end times. The mission end times are 1,000 to 10,000 hours, using increments of 1,000 hours.

Solution

The following figure shows the data as entered in Weibull++, as well as the calculated parameters.

The following figures show the probability plot with the 90% two-sided confidence bounds and the pdf plot.

Both the reliability and MTTF can be easily obtained from the QCP. The QCP, with results, for both cases is shown in the next two figures.

To obtain tabulated values for the failure rate, use the Analysis Workbook or General Spreadsheet features that are included in Weibull++. (For more information on these features, please refer to the Weibull++ User's Guide. For a step-by-step example on creating Weibull++ reports, please see the Quick Start Guide). The following worksheet shows the mission times and the corresponding failure rates.

Suspension Data Example

19 units are being reliability tested and the following is a table of their times-to-failure and suspensions.

Non-Grouped Data Times-to-Failure Data with Suspensions
Data point index	Last Inspected	State End Time
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.07 \\ & {{{\hat{\sigma }}}_{T}}= & 28.41. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 46.40 \\ & {{{\hat{\sigma }}}_{T}}= & 28.64. \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 47.34 \\ & {{{\hat{\sigma }}}_{T}}= & 29.96. \end{align}\,\! }[/math]

Interval Censored Data Example

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Interval Data
Data point index	Last Inspected	State End Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

This is a sequence of interval times-to-failure data. Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 7.740. \end{align}\,\! }[/math]

For rank regression on x:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 9.03. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on y (RRY) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.39 \\ & {{{\hat{\sigma }}}_{T}}= & 9.25. \end{align}\,\! }[/math]

The following plot shows the results if the data were analyzed using the rank regression on X (RRX) method.

Mixed Data Types Example

Suppose our data set includes left and right censored, interval censored and complete data, as shown in the following table.

Grouped Data Times-to-Failure with Suspensions and Intervals (Interval, Left and Right Censored)
Data point index	Number in State	Last Inspection	State (S or F)	State End Time
1	1	10	F	10
2	1	20	S	20
3	2	0	F	30
4	2	40	F	40
5	1	50	F	50
6	1	60	S	60
7	1	70	F	70
8	2	20	F	80
9	1	10	F	85
10	1	100	F	100

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.11 \\ & {{{\hat{\sigma }}}_{T}}= & 26.42 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 49.99 \\ & {{{\hat{\sigma }}}_{T}}= & 30.17 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 51.61 \\ & {{{\hat{\sigma }}}_{T}}= & 33.07 \end{align}\,\! }[/math]

Comparison of Analysis Methods

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Times-to-Failure Data
Data point index	State F or S	State End Time
1	F	2
2	F	5
3	F	11
4	F	23
5	F	29
6	F	37
7	F	43
8	F	59

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 18.57 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 21.64 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 22.28. \end{align}\,\! }[/math]

The Normal Distribution

The normal distribution, also known as the Gaussian distribution, is the most widely-used general purpose distribution. It is for this reason that it is included among the lifetime distributions commonly used for reliability and life data analysis. There are some who argue that the normal distribution is inappropriate for modeling lifetime data because the left-hand limit of the distribution extends to negative infinity. This could conceivably result in modeling negative times-to-failure. However, provided that the distribution in question has a relatively high mean and a relatively small standard deviation, the issue of negative failure times should not present itself as a problem. Nevertheless, the normal distribution has been shown to be useful for modeling the lifetimes of consumable items, such as printer toner cartridges.

Normal Probability Density Function

The [math]\displaystyle{ pdf }[/math] of the normal distribution is given by:

[math]\displaystyle{ f(t)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{t-\mu }{\sigma } \right)}^{2}}}} }[/math]

where:

[math]\displaystyle{ \mu }[/math] = mean of the normal times-to-faiure, also noted as [math]\displaystyle{ \bar{T} }[/math],

[math]\displaystyle{ \theta=\text{standard deviation of the times-to-failure} }[/math]

It is a two-parameter distribution with parameters [math]\displaystyle{ \mu }[/math] (or [math]\displaystyle{ \bar{T} }[/math] ) and [math]\displaystyle{ {{\sigma }} }[/math] , i.e. the mean and the standard deviation, respectively.

Normal Statistical Properties

The Normal Mean, Median and Mode

The normal mean or MTTF is actually one of the parameters of the distribution, usually denoted as [math]\displaystyle{ \mu . }[/math] Since the normal distribution is symmetrical, the median and the mode are always equal to the mean,

[math]\displaystyle{ \mu =\tilde{T}=\breve{T} }[/math]

The Normal Standard Deviation

As with the mean, the standard deviation for the normal distribution is actually one of the parameters, usually denoted as [math]\displaystyle{ {{\sigma }_{T}}. }[/math]

The Normal Reliability Function

The reliability for a mission of time [math]\displaystyle{ T }[/math] for the normal distribution is determined by:

[math]\displaystyle{ R(t)=\int_{t}^{\infty }f(x)dx=\int_{t}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx }[/math]

There is no closed-form solution for the normal reliability function. Solutions can be obtained via the use of standard normal tables. Since the application automatically solves for the reliability, we will not discuss manual solution methods. For interested readers, full explanations can be found in the references.

The Normal Conditional Reliability Function

The normal conditional reliability function is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{\int_{T+t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}{\int_{T}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Once again, the use of standard normal tables for the calculation of the normal conditional reliability is necessary, as there is no closed form solution.

The Normal Reliable Life

Since there is no closed-form solution for the normal reliability function, there will also be no closed-form solution for the normal reliable life. To determine the normal reliable life, one must solve:

[math]\displaystyle{ R(T)=\int_{T}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}dt }[/math]

for [math]\displaystyle{ T }[/math] .

The Normal Failure Rate Function

The instantaneous normal failure rate is given by:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}}{\int_{t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Characteristics of the Normal Distribution

Some of the specific characteristics of the normal distribution are the following:

• The normal [math]\displaystyle{ pdf }[/math] has a mean, [math]\displaystyle{ \bar{T} }[/math] , which is equal to the median, [math]\displaystyle{ \breve{T} }[/math], and also equal to the mode, [math]\displaystyle{ \tilde{T} }[/math], or [math]\displaystyle{ \bar{T}=\breve{T}=\tilde{T} }[/math]. This is because the normal distribution is symmetrical about its mean.

• The mean, [math]\displaystyle{ \mu }[/math] , or the mean life or the [math]\displaystyle{ MTTF }[/math] , is also the location parameter of the normal [math]\displaystyle{ pdf }[/math] , as it locates the [math]\displaystyle{ pdf }[/math] along the abscissa. It can assume values of [math]\displaystyle{ -\infty \lt \bar{T}\lt \infty }[/math] .
• The normal [math]\displaystyle{ pdf }[/math] has no shape parameter. This means that the normal [math]\displaystyle{ pdf }[/math] has only one shape, the bell shape, and this shape does not change.

• The standard deviation, [math]\displaystyle{ {{\sigma }} }[/math] , is the scale parameter of the normal [math]\displaystyle{ pdf }[/math] .

• As [math]\displaystyle{ {{\sigma }} }[/math] decreases, the [math]\displaystyle{ pdf }[/math] gets pushed toward the mean, or it becomes narrower and taller.

• As [math]\displaystyle{ {{\sigma }} }[/math] increases, the [math]\displaystyle{ pdf }[/math] spreads out away from the mean, or it becomes broader and shallower.

• The standard deviation can assume values of [math]\displaystyle{ 0\lt {{\sigma }}\lt \infty }[/math] .

• The greater the variability, the larger the value of [math]\displaystyle{ {{\sigma }} }[/math] , and vice versa.

• The standard deviation is also the distance between the mean and the point of inflection of the [math]\displaystyle{ pdf }[/math] , on each side of the mean. The point of inflection is that point of the [math]\displaystyle{ pdf }[/math] where the slope changes its value from a decreasing to an increasing one, or where the second derivative of the [math]\displaystyle{ pdf }[/math] has a value of zero.

• The normal [math]\displaystyle{ pdf }[/math] starts at [math]\displaystyle{ t=-\infty }[/math] with an [math]\displaystyle{ f(t)=0 }[/math] . As [math]\displaystyle{ t }[/math] increases, [math]\displaystyle{ f(t) }[/math] also increases, goes through its point of inflection and reaches its maximum value at [math]\displaystyle{ t=\bar{T} }[/math] . Thereafter, [math]\displaystyle{ f(t) }[/math] decreases, goes through its point of inflection, and assumes a value of [math]\displaystyle{ f(t)=0 }[/math] at [math]\displaystyle{ t=+\infty }[/math] .

Weibull++ Notes on Negative Time Values

One of the disadvantages of using the normal distribution for reliability calculations is the fact that the normal distribution starts at negative infinity. This can result in negative values for some of the results. Negative values for time are not accepted in most of the components of Weibull++, nor are they implemented. Certain components of the application reserve negative values for suspensions, or will not return negative results. For example, the Quick Calculation Pad will return a null value (zero) if the result is negative. Only the Free-Form (Probit) data sheet can accept negative values for the random variable (x-axis values).

Estimation of the Parameters

Probability Plotting

As described before, probability plotting involves plotting the failure times and associated unreliability estimates on specially constructed probability plotting paper. The form of this paper is based on a linearization of the [math]\displaystyle{ cdf }[/math] of the specific distribution. For the normal distribution, the cumulative density function can be written as:

[math]\displaystyle{ F(t)=\Phi \left( \frac{t-\mu }{{{\sigma }}} \right) }[/math]

or:

[math]\displaystyle{ {{\Phi }^{-1}}\left[ F(t) \right]=-\frac{\mu}{\sigma}+\frac{1}{\sigma}t }[/math]

where:

[math]\displaystyle{ \Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt }[/math]

Now, let:

[math]\displaystyle{ y={{\Phi }^{-1}}\left[ F(t) \right] }[/math]

[math]\displaystyle{ a=-\frac{\mu }{\sigma } }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\sigma } }[/math]

which results in the linear equation of:

[math]\displaystyle{ y=a+bT }[/math]

The normal probability paper resulting from this linearized [math]\displaystyle{ cdf }[/math] function is shown next.

Since the normal distribution is symmetrical, the area under the [math]\displaystyle{ pdf }[/math] curve from [math]\displaystyle{ -\infty }[/math] to [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ 0.5 }[/math] , as is the area from [math]\displaystyle{ \mu }[/math] to [math]\displaystyle{ +\infty }[/math] . Consequently, the value of [math]\displaystyle{ \mu }[/math] is said to be the point where [math]\displaystyle{ R(t)=Q(t)=50% }[/math] . This means that the estimate of [math]\displaystyle{ \mu }[/math] can be read from the point where the plotted line crosses the 50% unreliability line.

To determine the value of [math]\displaystyle{ \sigma }[/math] from the probability plot, it is first necessary to understand that the area under the [math]\displaystyle{ pdf }[/math] curve that lies between one standard deviation in either direction from the mean (or two standard deviations total) represents 68.3% of the area under the curve. This is represented graphically in the following figure.

Consequently, the interval between [math]\displaystyle{ Q(t)=84.15% }[/math] and [math]\displaystyle{ Q(t)=15.85% }[/math] represents two standard deviations, since this is an interval of 68.3% ( [math]\displaystyle{ 84.15-15.85=68.3 }[/math] ), and is centered on the mean at 50%. As a result, the standard deviation can be estimated from:

[math]\displaystyle{ \widehat{\sigma }=\frac{t(Q=84.15%)-t(Q=15.85%)}{2} }[/math]

That is: the value of [math]\displaystyle{ \widehat{\sigma } }[/math] is obtained by subtracting the time value where the plotted line crosses the 84.15% unreliability line from the time value where the plotted line crosses the 15.85% unreliability line and dividing the result by two. This process is illustrated in the following example.

Example 1:

Rank Regression on Y

Performing rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

The least squares parameter estimation method (regression analysis) was discussed in Parameter Estimation, and the following equations for regression on Y were derived:

[math]\displaystyle{ \begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\ =& \frac{\sum_{i=1}^N y_{i}}{N}-\hat{b}\frac{\sum_{i=1}^{N}x_{i}}{N}\\ \end{align} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}} }[/math]

In the case of the normal distribution, the equations for [math]\displaystyle{ {{y}_{i}} }[/math] and [math]\displaystyle{ {{x}_{i}} }[/math] are:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

where the values for [math]\displaystyle{ F({{T}_{i}}) }[/math] are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, [math]\displaystyle{ \widehat{\sigma } }[/math] and [math]\displaystyle{ \widehat{\mu } }[/math] can easily be obtained from above equations.

The Correlation Coefficient The estimator of the sample correlation coefficient, [math]\displaystyle{ \hat{\rho } }[/math] , is given by:

[math]\displaystyle{ \hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}} }[/math]

Example 2:

Rank Regression on X

As was mentioned previously, performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the fitted line is minimized.

Again, the first task is to bring our function, the probability of failure function for normal distribution, into a linear form. This step is exactly the same as in regression on Y analysis. All other equations apply in this case as they did for the regression on Y. The deviation from the previous analysis begins on the least squares fit step where: in this case, we treat [math]\displaystyle{ x }[/math] as the dependent variable and [math]\displaystyle{ y }[/math] as the independent variable. The best-fitting straight line for the data, for regression on X, is the straight line:

[math]\displaystyle{ x=\widehat{a}+\widehat{b}y }[/math]

The corresponding equations for [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are:

[math]\displaystyle{ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{N}} }[/math]

where:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

and the [math]\displaystyle{ F({{t}_{i}}) }[/math] values are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, solve the above linear equation for the unknown value of [math]\displaystyle{ y }[/math] which corresponds to:

[math]\displaystyle{ y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x }[/math]

Solving for the parameters, we get:

[math]\displaystyle{ a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{\mu }{\sigma }\Rightarrow \mu =\widehat{a} }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\widehat{b}}=\frac{1}{\sigma }\Rightarrow \sigma =\widehat{b} }[/math]

The correlation coefficient is evaluated as before.

Example 3:

The following examples illustrate the different types of life data that can be analyzed in Weibull++ using the normal distribution. For more information on the different types of life data, see Life Data Classification.

Complete Data Example

6 units are tested to failure. The following failure times data are obtained: 12125, 11260, 12080, 12825, 13550 and 14670 hours. Assuming that the data are normally distributed, do the following:

Objectives

1. Find the parameters for the data set, using the Rank Regression on X (RRX) parameter estimation method

2. Obtain the probability plot for the data with 90%, two-sided Type 1 confidence bounds.

3. Obtain the pdf plot for the data.

4. Using the Quick Calculation Pad (QCP), determine the reliability for a mission of 11,000 hours, as well as the upper and lower two-sided 90% confidence limit on this reliability.

5. Using the QCP, determine the MTTF, as well as the upper and lower two-sided 90% confidence limit on this MTTF.

6. Obtain tabulated values for the failure rate for 10 different mission end times. The mission end times are 1,000 to 10,000 hours, using increments of 1,000 hours.

Solution

The following figure shows the data as entered in Weibull++, as well as the calculated parameters.

The following figures show the probability plot with the 90% two-sided confidence bounds and the pdf plot.

Both the reliability and MTTF can be easily obtained from the QCP. The QCP, with results, for both cases is shown in the next two figures.

To obtain tabulated values for the failure rate, use the Analysis Workbook or General Spreadsheet features that are included in Weibull++. (For more information on these features, please refer to the Weibull++ User's Guide. For a step-by-step example on creating Weibull++ reports, please see the Quick Start Guide). The following worksheet shows the mission times and the corresponding failure rates.

Suspension Data Example

19 units are being reliability tested and the following is a table of their times-to-failure and suspensions.

Non-Grouped Data Times-to-Failure Data with Suspensions
Data point index	Last Inspected	State End Time
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.07 \\ & {{{\hat{\sigma }}}_{T}}= & 28.41. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 46.40 \\ & {{{\hat{\sigma }}}_{T}}= & 28.64. \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 47.34 \\ & {{{\hat{\sigma }}}_{T}}= & 29.96. \end{align}\,\! }[/math]

Interval Censored Data Example

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Interval Data
Data point index	Last Inspected	State End Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

This is a sequence of interval times-to-failure data. Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 7.740. \end{align}\,\! }[/math]

For rank regression on x:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 9.03. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on y (RRY) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.39 \\ & {{{\hat{\sigma }}}_{T}}= & 9.25. \end{align}\,\! }[/math]

The following plot shows the results if the data were analyzed using the rank regression on X (RRX) method.

Mixed Data Types Example

Suppose our data set includes left and right censored, interval censored and complete data, as shown in the following table.

Grouped Data Times-to-Failure with Suspensions and Intervals (Interval, Left and Right Censored)
Data point index	Number in State	Last Inspection	State (S or F)	State End Time
1	1	10	F	10
2	1	20	S	20
3	2	0	F	30
4	2	40	F	40
5	1	50	F	50
6	1	60	S	60
7	1	70	F	70
8	2	20	F	80
9	1	10	F	85
10	1	100	F	100

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.11 \\ & {{{\hat{\sigma }}}_{T}}= & 26.42 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 49.99 \\ & {{{\hat{\sigma }}}_{T}}= & 30.17 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 51.61 \\ & {{{\hat{\sigma }}}_{T}}= & 33.07 \end{align}\,\! }[/math]

Comparison of Analysis Methods

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Times-to-Failure Data
Data point index	State F or S	State End Time
1	F	2
2	F	5
3	F	11
4	F	23
5	F	29
6	F	37
7	F	43
8	F	59

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 18.57 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 21.64 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 22.28. \end{align}\,\! }[/math]

The Normal Distribution

The normal distribution, also known as the Gaussian distribution, is the most widely-used general purpose distribution. It is for this reason that it is included among the lifetime distributions commonly used for reliability and life data analysis. There are some who argue that the normal distribution is inappropriate for modeling lifetime data because the left-hand limit of the distribution extends to negative infinity. This could conceivably result in modeling negative times-to-failure. However, provided that the distribution in question has a relatively high mean and a relatively small standard deviation, the issue of negative failure times should not present itself as a problem. Nevertheless, the normal distribution has been shown to be useful for modeling the lifetimes of consumable items, such as printer toner cartridges.

Normal Probability Density Function

The [math]\displaystyle{ pdf }[/math] of the normal distribution is given by:

[math]\displaystyle{ f(t)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{t-\mu }{\sigma } \right)}^{2}}}} }[/math]

where:

[math]\displaystyle{ \mu }[/math] = mean of the normal times-to-faiure, also noted as [math]\displaystyle{ \bar{T} }[/math],

[math]\displaystyle{ \theta=\text{standard deviation of the times-to-failure} }[/math]

It is a two-parameter distribution with parameters [math]\displaystyle{ \mu }[/math] (or [math]\displaystyle{ \bar{T} }[/math] ) and [math]\displaystyle{ {{\sigma }} }[/math] , i.e. the mean and the standard deviation, respectively.

Normal Statistical Properties

The Normal Mean, Median and Mode

The normal mean or MTTF is actually one of the parameters of the distribution, usually denoted as [math]\displaystyle{ \mu . }[/math] Since the normal distribution is symmetrical, the median and the mode are always equal to the mean,

[math]\displaystyle{ \mu =\tilde{T}=\breve{T} }[/math]

The Normal Standard Deviation

As with the mean, the standard deviation for the normal distribution is actually one of the parameters, usually denoted as [math]\displaystyle{ {{\sigma }_{T}}. }[/math]

The Normal Reliability Function

The reliability for a mission of time [math]\displaystyle{ T }[/math] for the normal distribution is determined by:

[math]\displaystyle{ R(t)=\int_{t}^{\infty }f(x)dx=\int_{t}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx }[/math]

There is no closed-form solution for the normal reliability function. Solutions can be obtained via the use of standard normal tables. Since the application automatically solves for the reliability, we will not discuss manual solution methods. For interested readers, full explanations can be found in the references.

The Normal Conditional Reliability Function

The normal conditional reliability function is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{\int_{T+t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}{\int_{T}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Once again, the use of standard normal tables for the calculation of the normal conditional reliability is necessary, as there is no closed form solution.

The Normal Reliable Life

Since there is no closed-form solution for the normal reliability function, there will also be no closed-form solution for the normal reliable life. To determine the normal reliable life, one must solve:

[math]\displaystyle{ R(T)=\int_{T}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}dt }[/math]

for [math]\displaystyle{ T }[/math] .

The Normal Failure Rate Function

The instantaneous normal failure rate is given by:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}}{\int_{t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Characteristics of the Normal Distribution

Some of the specific characteristics of the normal distribution are the following:

• The normal [math]\displaystyle{ pdf }[/math] has a mean, [math]\displaystyle{ \bar{T} }[/math] , which is equal to the median, [math]\displaystyle{ \breve{T} }[/math], and also equal to the mode, [math]\displaystyle{ \tilde{T} }[/math], or [math]\displaystyle{ \bar{T}=\breve{T}=\tilde{T} }[/math]. This is because the normal distribution is symmetrical about its mean.

• The mean, [math]\displaystyle{ \mu }[/math] , or the mean life or the [math]\displaystyle{ MTTF }[/math] , is also the location parameter of the normal [math]\displaystyle{ pdf }[/math] , as it locates the [math]\displaystyle{ pdf }[/math] along the abscissa. It can assume values of [math]\displaystyle{ -\infty \lt \bar{T}\lt \infty }[/math] .
• The normal [math]\displaystyle{ pdf }[/math] has no shape parameter. This means that the normal [math]\displaystyle{ pdf }[/math] has only one shape, the bell shape, and this shape does not change.

• The standard deviation, [math]\displaystyle{ {{\sigma }} }[/math] , is the scale parameter of the normal [math]\displaystyle{ pdf }[/math] .

• As [math]\displaystyle{ {{\sigma }} }[/math] decreases, the [math]\displaystyle{ pdf }[/math] gets pushed toward the mean, or it becomes narrower and taller.

• As [math]\displaystyle{ {{\sigma }} }[/math] increases, the [math]\displaystyle{ pdf }[/math] spreads out away from the mean, or it becomes broader and shallower.

• The standard deviation can assume values of [math]\displaystyle{ 0\lt {{\sigma }}\lt \infty }[/math] .

• The greater the variability, the larger the value of [math]\displaystyle{ {{\sigma }} }[/math] , and vice versa.

• The standard deviation is also the distance between the mean and the point of inflection of the [math]\displaystyle{ pdf }[/math] , on each side of the mean. The point of inflection is that point of the [math]\displaystyle{ pdf }[/math] where the slope changes its value from a decreasing to an increasing one, or where the second derivative of the [math]\displaystyle{ pdf }[/math] has a value of zero.

• The normal [math]\displaystyle{ pdf }[/math] starts at [math]\displaystyle{ t=-\infty }[/math] with an [math]\displaystyle{ f(t)=0 }[/math] . As [math]\displaystyle{ t }[/math] increases, [math]\displaystyle{ f(t) }[/math] also increases, goes through its point of inflection and reaches its maximum value at [math]\displaystyle{ t=\bar{T} }[/math] . Thereafter, [math]\displaystyle{ f(t) }[/math] decreases, goes through its point of inflection, and assumes a value of [math]\displaystyle{ f(t)=0 }[/math] at [math]\displaystyle{ t=+\infty }[/math] .

Weibull++ Notes on Negative Time Values

One of the disadvantages of using the normal distribution for reliability calculations is the fact that the normal distribution starts at negative infinity. This can result in negative values for some of the results. Negative values for time are not accepted in most of the components of Weibull++, nor are they implemented. Certain components of the application reserve negative values for suspensions, or will not return negative results. For example, the Quick Calculation Pad will return a null value (zero) if the result is negative. Only the Free-Form (Probit) data sheet can accept negative values for the random variable (x-axis values).

Estimation of the Parameters

Probability Plotting

As described before, probability plotting involves plotting the failure times and associated unreliability estimates on specially constructed probability plotting paper. The form of this paper is based on a linearization of the [math]\displaystyle{ cdf }[/math] of the specific distribution. For the normal distribution, the cumulative density function can be written as:

[math]\displaystyle{ F(t)=\Phi \left( \frac{t-\mu }{{{\sigma }}} \right) }[/math]

or:

[math]\displaystyle{ {{\Phi }^{-1}}\left[ F(t) \right]=-\frac{\mu}{\sigma}+\frac{1}{\sigma}t }[/math]

where:

[math]\displaystyle{ \Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt }[/math]

Now, let:

[math]\displaystyle{ y={{\Phi }^{-1}}\left[ F(t) \right] }[/math]

[math]\displaystyle{ a=-\frac{\mu }{\sigma } }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\sigma } }[/math]

which results in the linear equation of:

[math]\displaystyle{ y=a+bT }[/math]

The normal probability paper resulting from this linearized [math]\displaystyle{ cdf }[/math] function is shown next.

Since the normal distribution is symmetrical, the area under the [math]\displaystyle{ pdf }[/math] curve from [math]\displaystyle{ -\infty }[/math] to [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ 0.5 }[/math] , as is the area from [math]\displaystyle{ \mu }[/math] to [math]\displaystyle{ +\infty }[/math] . Consequently, the value of [math]\displaystyle{ \mu }[/math] is said to be the point where [math]\displaystyle{ R(t)=Q(t)=50% }[/math] . This means that the estimate of [math]\displaystyle{ \mu }[/math] can be read from the point where the plotted line crosses the 50% unreliability line.

To determine the value of [math]\displaystyle{ \sigma }[/math] from the probability plot, it is first necessary to understand that the area under the [math]\displaystyle{ pdf }[/math] curve that lies between one standard deviation in either direction from the mean (or two standard deviations total) represents 68.3% of the area under the curve. This is represented graphically in the following figure.

Consequently, the interval between [math]\displaystyle{ Q(t)=84.15% }[/math] and [math]\displaystyle{ Q(t)=15.85% }[/math] represents two standard deviations, since this is an interval of 68.3% ( [math]\displaystyle{ 84.15-15.85=68.3 }[/math] ), and is centered on the mean at 50%. As a result, the standard deviation can be estimated from:

[math]\displaystyle{ \widehat{\sigma }=\frac{t(Q=84.15%)-t(Q=15.85%)}{2} }[/math]

That is: the value of [math]\displaystyle{ \widehat{\sigma } }[/math] is obtained by subtracting the time value where the plotted line crosses the 84.15% unreliability line from the time value where the plotted line crosses the 15.85% unreliability line and dividing the result by two. This process is illustrated in the following example.

Example 1:

Rank Regression on Y

Performing rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

The least squares parameter estimation method (regression analysis) was discussed in Parameter Estimation, and the following equations for regression on Y were derived:

[math]\displaystyle{ \begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\ =& \frac{\sum_{i=1}^N y_{i}}{N}-\hat{b}\frac{\sum_{i=1}^{N}x_{i}}{N}\\ \end{align} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}} }[/math]

In the case of the normal distribution, the equations for [math]\displaystyle{ {{y}_{i}} }[/math] and [math]\displaystyle{ {{x}_{i}} }[/math] are:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

where the values for [math]\displaystyle{ F({{T}_{i}}) }[/math] are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, [math]\displaystyle{ \widehat{\sigma } }[/math] and [math]\displaystyle{ \widehat{\mu } }[/math] can easily be obtained from above equations.

The Correlation Coefficient The estimator of the sample correlation coefficient, [math]\displaystyle{ \hat{\rho } }[/math] , is given by:

[math]\displaystyle{ \hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}} }[/math]

Example 2:

Rank Regression on X

As was mentioned previously, performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the fitted line is minimized.

Again, the first task is to bring our function, the probability of failure function for normal distribution, into a linear form. This step is exactly the same as in regression on Y analysis. All other equations apply in this case as they did for the regression on Y. The deviation from the previous analysis begins on the least squares fit step where: in this case, we treat [math]\displaystyle{ x }[/math] as the dependent variable and [math]\displaystyle{ y }[/math] as the independent variable. The best-fitting straight line for the data, for regression on X, is the straight line:

[math]\displaystyle{ x=\widehat{a}+\widehat{b}y }[/math]

The corresponding equations for [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are:

[math]\displaystyle{ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{N}} }[/math]

where:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

and the [math]\displaystyle{ F({{t}_{i}}) }[/math] values are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, solve the above linear equation for the unknown value of [math]\displaystyle{ y }[/math] which corresponds to:

[math]\displaystyle{ y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x }[/math]

Solving for the parameters, we get:

[math]\displaystyle{ a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{\mu }{\sigma }\Rightarrow \mu =\widehat{a} }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\widehat{b}}=\frac{1}{\sigma }\Rightarrow \sigma =\widehat{b} }[/math]

The correlation coefficient is evaluated as before.

Example 3:

The following examples illustrate the different types of life data that can be analyzed in Weibull++ using the normal distribution. For more information on the different types of life data, see Life Data Classification.

Complete Data Example

6 units are tested to failure. The following failure times data are obtained: 12125, 11260, 12080, 12825, 13550 and 14670 hours. Assuming that the data are normally distributed, do the following:

Objectives

1. Find the parameters for the data set, using the Rank Regression on X (RRX) parameter estimation method

2. Obtain the probability plot for the data with 90%, two-sided Type 1 confidence bounds.

3. Obtain the pdf plot for the data.

4. Using the Quick Calculation Pad (QCP), determine the reliability for a mission of 11,000 hours, as well as the upper and lower two-sided 90% confidence limit on this reliability.

5. Using the QCP, determine the MTTF, as well as the upper and lower two-sided 90% confidence limit on this MTTF.

6. Obtain tabulated values for the failure rate for 10 different mission end times. The mission end times are 1,000 to 10,000 hours, using increments of 1,000 hours.

Solution

The following figure shows the data as entered in Weibull++, as well as the calculated parameters.

The following figures show the probability plot with the 90% two-sided confidence bounds and the pdf plot.

Both the reliability and MTTF can be easily obtained from the QCP. The QCP, with results, for both cases is shown in the next two figures.

To obtain tabulated values for the failure rate, use the Analysis Workbook or General Spreadsheet features that are included in Weibull++. (For more information on these features, please refer to the Weibull++ User's Guide. For a step-by-step example on creating Weibull++ reports, please see the Quick Start Guide). The following worksheet shows the mission times and the corresponding failure rates.

Suspension Data Example

19 units are being reliability tested and the following is a table of their times-to-failure and suspensions.

Non-Grouped Data Times-to-Failure Data with Suspensions
Data point index	Last Inspected	State End Time
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.07 \\ & {{{\hat{\sigma }}}_{T}}= & 28.41. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 46.40 \\ & {{{\hat{\sigma }}}_{T}}= & 28.64. \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 47.34 \\ & {{{\hat{\sigma }}}_{T}}= & 29.96. \end{align}\,\! }[/math]

Interval Censored Data Example

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Interval Data
Data point index	Last Inspected	State End Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

This is a sequence of interval times-to-failure data. Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 7.740. \end{align}\,\! }[/math]

For rank regression on x:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 9.03. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on y (RRY) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.39 \\ & {{{\hat{\sigma }}}_{T}}= & 9.25. \end{align}\,\! }[/math]

The following plot shows the results if the data were analyzed using the rank regression on X (RRX) method.

Mixed Data Types Example

Suppose our data set includes left and right censored, interval censored and complete data, as shown in the following table.

Grouped Data Times-to-Failure with Suspensions and Intervals (Interval, Left and Right Censored)
Data point index	Number in State	Last Inspection	State (S or F)	State End Time
1	1	10	F	10
2	1	20	S	20
3	2	0	F	30
4	2	40	F	40
5	1	50	F	50
6	1	60	S	60
7	1	70	F	70
8	2	20	F	80
9	1	10	F	85
10	1	100	F	100

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.11 \\ & {{{\hat{\sigma }}}_{T}}= & 26.42 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 49.99 \\ & {{{\hat{\sigma }}}_{T}}= & 30.17 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 51.61 \\ & {{{\hat{\sigma }}}_{T}}= & 33.07 \end{align}\,\! }[/math]

Comparison of Analysis Methods

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Times-to-Failure Data
Data point index	State F or S	State End Time
1	F	2
2	F	5
3	F	11
4	F	23
5	F	29
6	F	37
7	F	43
8	F	59

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 18.57 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 21.64 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 22.28. \end{align}\,\! }[/math]

The Normal Distribution

The normal distribution, also known as the Gaussian distribution, is the most widely-used general purpose distribution. It is for this reason that it is included among the lifetime distributions commonly used for reliability and life data analysis. There are some who argue that the normal distribution is inappropriate for modeling lifetime data because the left-hand limit of the distribution extends to negative infinity. This could conceivably result in modeling negative times-to-failure. However, provided that the distribution in question has a relatively high mean and a relatively small standard deviation, the issue of negative failure times should not present itself as a problem. Nevertheless, the normal distribution has been shown to be useful for modeling the lifetimes of consumable items, such as printer toner cartridges.

Normal Probability Density Function

The [math]\displaystyle{ pdf }[/math] of the normal distribution is given by:

[math]\displaystyle{ f(t)=\frac{1}{\sigma \sqrt{2\pi }}{{e}^{-\frac{1}{2}{{\left( \frac{t-\mu }{\sigma } \right)}^{2}}}} }[/math]

where:

[math]\displaystyle{ \mu }[/math] = mean of the normal times-to-faiure, also noted as [math]\displaystyle{ \bar{T} }[/math],

[math]\displaystyle{ \theta=\text{standard deviation of the times-to-failure} }[/math]

It is a two-parameter distribution with parameters [math]\displaystyle{ \mu }[/math] (or [math]\displaystyle{ \bar{T} }[/math] ) and [math]\displaystyle{ {{\sigma }} }[/math] , i.e. the mean and the standard deviation, respectively.

Normal Statistical Properties

The Normal Mean, Median and Mode

The normal mean or MTTF is actually one of the parameters of the distribution, usually denoted as [math]\displaystyle{ \mu . }[/math] Since the normal distribution is symmetrical, the median and the mode are always equal to the mean,

[math]\displaystyle{ \mu =\tilde{T}=\breve{T} }[/math]

The Normal Standard Deviation

As with the mean, the standard deviation for the normal distribution is actually one of the parameters, usually denoted as [math]\displaystyle{ {{\sigma }_{T}}. }[/math]

The Normal Reliability Function

The reliability for a mission of time [math]\displaystyle{ T }[/math] for the normal distribution is determined by:

[math]\displaystyle{ R(t)=\int_{t}^{\infty }f(x)dx=\int_{t}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx }[/math]

There is no closed-form solution for the normal reliability function. Solutions can be obtained via the use of standard normal tables. Since the application automatically solves for the reliability, we will not discuss manual solution methods. For interested readers, full explanations can be found in the references.

The Normal Conditional Reliability Function

The normal conditional reliability function is given by:

[math]\displaystyle{ R(t|T)=\frac{R(T+t)}{R(T)}=\frac{\int_{T+t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx}{\int_{T}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Once again, the use of standard normal tables for the calculation of the normal conditional reliability is necessary, as there is no closed form solution.

The Normal Reliable Life

Since there is no closed-form solution for the normal reliability function, there will also be no closed-form solution for the normal reliable life. To determine the normal reliable life, one must solve:

[math]\displaystyle{ R(T)=\int_{T}^{\infty }\frac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}dt }[/math]

for [math]\displaystyle{ T }[/math] .

The Normal Failure Rate Function

The instantaneous normal failure rate is given by:

[math]\displaystyle{ \lambda (t)=\frac{f(t)}{R(t)}=\frac{\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{t-\mu }{{{\sigma }}} \right)}^{2}}}}}{\int_{t}^{\infty }\tfrac{1}{{{\sigma }}\sqrt{2\pi }}{{e}^{-\tfrac{1}{2}{{\left( \tfrac{x-\mu }{{{\sigma }}} \right)}^{2}}}}dx} }[/math]

Characteristics of the Normal Distribution

Some of the specific characteristics of the normal distribution are the following:

• The normal [math]\displaystyle{ pdf }[/math] has a mean, [math]\displaystyle{ \bar{T} }[/math] , which is equal to the median, [math]\displaystyle{ \breve{T} }[/math], and also equal to the mode, [math]\displaystyle{ \tilde{T} }[/math], or [math]\displaystyle{ \bar{T}=\breve{T}=\tilde{T} }[/math]. This is because the normal distribution is symmetrical about its mean.

• The mean, [math]\displaystyle{ \mu }[/math] , or the mean life or the [math]\displaystyle{ MTTF }[/math] , is also the location parameter of the normal [math]\displaystyle{ pdf }[/math] , as it locates the [math]\displaystyle{ pdf }[/math] along the abscissa. It can assume values of [math]\displaystyle{ -\infty \lt \bar{T}\lt \infty }[/math] .
• The normal [math]\displaystyle{ pdf }[/math] has no shape parameter. This means that the normal [math]\displaystyle{ pdf }[/math] has only one shape, the bell shape, and this shape does not change.

• The standard deviation, [math]\displaystyle{ {{\sigma }} }[/math] , is the scale parameter of the normal [math]\displaystyle{ pdf }[/math] .

• As [math]\displaystyle{ {{\sigma }} }[/math] decreases, the [math]\displaystyle{ pdf }[/math] gets pushed toward the mean, or it becomes narrower and taller.

• As [math]\displaystyle{ {{\sigma }} }[/math] increases, the [math]\displaystyle{ pdf }[/math] spreads out away from the mean, or it becomes broader and shallower.

• The standard deviation can assume values of [math]\displaystyle{ 0\lt {{\sigma }}\lt \infty }[/math] .

• The greater the variability, the larger the value of [math]\displaystyle{ {{\sigma }} }[/math] , and vice versa.

• The standard deviation is also the distance between the mean and the point of inflection of the [math]\displaystyle{ pdf }[/math] , on each side of the mean. The point of inflection is that point of the [math]\displaystyle{ pdf }[/math] where the slope changes its value from a decreasing to an increasing one, or where the second derivative of the [math]\displaystyle{ pdf }[/math] has a value of zero.

• The normal [math]\displaystyle{ pdf }[/math] starts at [math]\displaystyle{ t=-\infty }[/math] with an [math]\displaystyle{ f(t)=0 }[/math] . As [math]\displaystyle{ t }[/math] increases, [math]\displaystyle{ f(t) }[/math] also increases, goes through its point of inflection and reaches its maximum value at [math]\displaystyle{ t=\bar{T} }[/math] . Thereafter, [math]\displaystyle{ f(t) }[/math] decreases, goes through its point of inflection, and assumes a value of [math]\displaystyle{ f(t)=0 }[/math] at [math]\displaystyle{ t=+\infty }[/math] .

Weibull++ Notes on Negative Time Values

One of the disadvantages of using the normal distribution for reliability calculations is the fact that the normal distribution starts at negative infinity. This can result in negative values for some of the results. Negative values for time are not accepted in most of the components of Weibull++, nor are they implemented. Certain components of the application reserve negative values for suspensions, or will not return negative results. For example, the Quick Calculation Pad will return a null value (zero) if the result is negative. Only the Free-Form (Probit) data sheet can accept negative values for the random variable (x-axis values).

Estimation of the Parameters

Probability Plotting

As described before, probability plotting involves plotting the failure times and associated unreliability estimates on specially constructed probability plotting paper. The form of this paper is based on a linearization of the [math]\displaystyle{ cdf }[/math] of the specific distribution. For the normal distribution, the cumulative density function can be written as:

[math]\displaystyle{ F(t)=\Phi \left( \frac{t-\mu }{{{\sigma }}} \right) }[/math]

or:

[math]\displaystyle{ {{\Phi }^{-1}}\left[ F(t) \right]=-\frac{\mu}{\sigma}+\frac{1}{\sigma}t }[/math]

where:

[math]\displaystyle{ \Phi (x)=\frac{1}{\sqrt{2\pi }}\int_{-\infty }^{x}{{e}^{-\tfrac{{{t}^{2}}}{2}}}dt }[/math]

Now, let:

[math]\displaystyle{ y={{\Phi }^{-1}}\left[ F(t) \right] }[/math]

[math]\displaystyle{ a=-\frac{\mu }{\sigma } }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\sigma } }[/math]

which results in the linear equation of:

[math]\displaystyle{ y=a+bT }[/math]

The normal probability paper resulting from this linearized [math]\displaystyle{ cdf }[/math] function is shown next.

Since the normal distribution is symmetrical, the area under the [math]\displaystyle{ pdf }[/math] curve from [math]\displaystyle{ -\infty }[/math] to [math]\displaystyle{ \mu }[/math] is [math]\displaystyle{ 0.5 }[/math] , as is the area from [math]\displaystyle{ \mu }[/math] to [math]\displaystyle{ +\infty }[/math] . Consequently, the value of [math]\displaystyle{ \mu }[/math] is said to be the point where [math]\displaystyle{ R(t)=Q(t)=50% }[/math] . This means that the estimate of [math]\displaystyle{ \mu }[/math] can be read from the point where the plotted line crosses the 50% unreliability line.

To determine the value of [math]\displaystyle{ \sigma }[/math] from the probability plot, it is first necessary to understand that the area under the [math]\displaystyle{ pdf }[/math] curve that lies between one standard deviation in either direction from the mean (or two standard deviations total) represents 68.3% of the area under the curve. This is represented graphically in the following figure.

Consequently, the interval between [math]\displaystyle{ Q(t)=84.15% }[/math] and [math]\displaystyle{ Q(t)=15.85% }[/math] represents two standard deviations, since this is an interval of 68.3% ( [math]\displaystyle{ 84.15-15.85=68.3 }[/math] ), and is centered on the mean at 50%. As a result, the standard deviation can be estimated from:

[math]\displaystyle{ \widehat{\sigma }=\frac{t(Q=84.15%)-t(Q=15.85%)}{2} }[/math]

That is: the value of [math]\displaystyle{ \widehat{\sigma } }[/math] is obtained by subtracting the time value where the plotted line crosses the 84.15% unreliability line from the time value where the plotted line crosses the 15.85% unreliability line and dividing the result by two. This process is illustrated in the following example.

Example 1:

Rank Regression on Y

Performing rank regression on Y requires that a straight line be fitted to a set of data points such that the sum of the squares of the vertical deviations from the points to the line is minimized.

The least squares parameter estimation method (regression analysis) was discussed in Parameter Estimation, and the following equations for regression on Y were derived:

[math]\displaystyle{ \begin{align}\hat{a}= & \bar{b}-\hat{b}\bar{x} \\ =& \frac{\sum_{i=1}^N y_{i}}{N}-\hat{b}\frac{\sum_{i=1}^{N}x_{i}}{N}\\ \end{align} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,x_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}} \right)}^{2}}}{N}} }[/math]

In the case of the normal distribution, the equations for [math]\displaystyle{ {{y}_{i}} }[/math] and [math]\displaystyle{ {{x}_{i}} }[/math] are:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

where the values for [math]\displaystyle{ F({{T}_{i}}) }[/math] are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, [math]\displaystyle{ \widehat{\sigma } }[/math] and [math]\displaystyle{ \widehat{\mu } }[/math] can easily be obtained from above equations.

The Correlation Coefficient The estimator of the sample correlation coefficient, [math]\displaystyle{ \hat{\rho } }[/math] , is given by:

[math]\displaystyle{ \hat{\rho }=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,({{x}_{i}}-\overline{x})({{y}_{i}}-\overline{y})}{\sqrt{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{x}_{i}}-\overline{x})}^{2}}\cdot \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{({{y}_{i}}-\overline{y})}^{2}}}} }[/math]

Example 2:

Rank Regression on X

As was mentioned previously, performing a rank regression on X requires that a straight line be fitted to a set of data points such that the sum of the squares of the horizontal deviations from the points to the fitted line is minimized.

Again, the first task is to bring our function, the probability of failure function for normal distribution, into a linear form. This step is exactly the same as in regression on Y analysis. All other equations apply in this case as they did for the regression on Y. The deviation from the previous analysis begins on the least squares fit step where: in this case, we treat [math]\displaystyle{ x }[/math] as the dependent variable and [math]\displaystyle{ y }[/math] as the independent variable. The best-fitting straight line for the data, for regression on X, is the straight line:

[math]\displaystyle{ x=\widehat{a}+\widehat{b}y }[/math]

The corresponding equations for [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are:

[math]\displaystyle{ \hat{a}=\overline{x}-\hat{b}\overline{y}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}}{N}-\hat{b}\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N} }[/math]

and:

[math]\displaystyle{ \hat{b}=\frac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}{{y}_{i}}-\tfrac{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{x}_{i}}\underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}}}{N}}{\underset{i=1}{\overset{N}{\mathop{\sum }}}\,y_{i}^{2}-\tfrac{{{\left( \underset{i=1}{\overset{N}{\mathop{\sum }}}\,{{y}_{i}} \right)}^{2}}}{N}} }[/math]

where:

[math]\displaystyle{ {{y}_{i}}={{\Phi }^{-1}}\left[ F({{t}_{i}}) \right] }[/math]

and:

[math]\displaystyle{ {{x}_{i}}={{t}_{i}} }[/math]

and the [math]\displaystyle{ F({{t}_{i}}) }[/math] values are estimated from the median ranks. Once [math]\displaystyle{ \widehat{a} }[/math] and [math]\displaystyle{ \widehat{b} }[/math] are obtained, solve the above linear equation for the unknown value of [math]\displaystyle{ y }[/math] which corresponds to:

[math]\displaystyle{ y=-\frac{\widehat{a}}{\widehat{b}}+\frac{1}{\widehat{b}}x }[/math]

Solving for the parameters, we get:

[math]\displaystyle{ a=-\frac{\widehat{a}}{\widehat{b}}=-\frac{\mu }{\sigma }\Rightarrow \mu =\widehat{a} }[/math]

and:

[math]\displaystyle{ b=\frac{1}{\widehat{b}}=\frac{1}{\sigma }\Rightarrow \sigma =\widehat{b} }[/math]

The correlation coefficient is evaluated as before.

Example 3:

The following examples illustrate the different types of life data that can be analyzed in Weibull++ using the normal distribution. For more information on the different types of life data, see Life Data Classification.

Complete Data Example

6 units are tested to failure. The following failure times data are obtained: 12125, 11260, 12080, 12825, 13550 and 14670 hours. Assuming that the data are normally distributed, do the following:

Objectives

1. Find the parameters for the data set, using the Rank Regression on X (RRX) parameter estimation method

2. Obtain the probability plot for the data with 90%, two-sided Type 1 confidence bounds.

3. Obtain the pdf plot for the data.

4. Using the Quick Calculation Pad (QCP), determine the reliability for a mission of 11,000 hours, as well as the upper and lower two-sided 90% confidence limit on this reliability.

5. Using the QCP, determine the MTTF, as well as the upper and lower two-sided 90% confidence limit on this MTTF.

6. Obtain tabulated values for the failure rate for 10 different mission end times. The mission end times are 1,000 to 10,000 hours, using increments of 1,000 hours.

Solution

The following figure shows the data as entered in Weibull++, as well as the calculated parameters.

The following figures show the probability plot with the 90% two-sided confidence bounds and the pdf plot.

Both the reliability and MTTF can be easily obtained from the QCP. The QCP, with results, for both cases is shown in the next two figures.

To obtain tabulated values for the failure rate, use the Analysis Workbook or General Spreadsheet features that are included in Weibull++. (For more information on these features, please refer to the Weibull++ User's Guide. For a step-by-step example on creating Weibull++ reports, please see the Quick Start Guide). The following worksheet shows the mission times and the corresponding failure rates.

Suspension Data Example

19 units are being reliability tested and the following is a table of their times-to-failure and suspensions.

Non-Grouped Data Times-to-Failure Data with Suspensions
Data point index	Last Inspected	State End Time
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.07 \\ & {{{\hat{\sigma }}}_{T}}= & 28.41. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 46.40 \\ & {{{\hat{\sigma }}}_{T}}= & 28.64. \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 47.34 \\ & {{{\hat{\sigma }}}_{T}}= & 29.96. \end{align}\,\! }[/math]

Interval Censored Data Example

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Interval Data
Data point index	Last Inspected	State End Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

This is a sequence of interval times-to-failure data. Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 7.740. \end{align}\,\! }[/math]

For rank regression on x:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 9.03. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on y (RRY) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.39 \\ & {{{\hat{\sigma }}}_{T}}= & 9.25. \end{align}\,\! }[/math]

The following plot shows the results if the data were analyzed using the rank regression on X (RRX) method.

Mixed Data Types Example

Suppose our data set includes left and right censored, interval censored and complete data, as shown in the following table.

Grouped Data Times-to-Failure with Suspensions and Intervals (Interval, Left and Right Censored)
Data point index	Number in State	Last Inspection	State (S or F)	State End Time
1	1	10	F	10
2	1	20	S	20
3	2	0	F	30
4	2	40	F	40
5	1	50	F	50
6	1	60	S	60
7	1	70	F	70
8	2	20	F	80
9	1	10	F	85
10	1	100	F	100

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.11 \\ & {{{\hat{\sigma }}}_{T}}= & 26.42 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 49.99 \\ & {{{\hat{\sigma }}}_{T}}= & 30.17 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 51.61 \\ & {{{\hat{\sigma }}}_{T}}= & 33.07 \end{align}\,\! }[/math]

Comparison of Analysis Methods

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Times-to-Failure Data
Data point index	State F or S	State End Time
1	F	2
2	F	5
3	F	11
4	F	23
5	F	29
6	F	37
7	F	43
8	F	59

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 18.57 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 21.64 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 22.28. \end{align}\,\! }[/math]

The following examples illustrate the different types of life data that can be analyzed in Weibull++ using the normal distribution. For more information on the different types of life data, see Life Data Classification.

Complete Data Example

6 units are tested to failure. The following failure times data are obtained: 12125, 11260, 12080, 12825, 13550 and 14670 hours. Assuming that the data are normally distributed, do the following:

Objectives

1. Find the parameters for the data set, using the Rank Regression on X (RRX) parameter estimation method

2. Obtain the probability plot for the data with 90%, two-sided Type 1 confidence bounds.

3. Obtain the pdf plot for the data.

4. Using the Quick Calculation Pad (QCP), determine the reliability for a mission of 11,000 hours, as well as the upper and lower two-sided 90% confidence limit on this reliability.

5. Using the QCP, determine the MTTF, as well as the upper and lower two-sided 90% confidence limit on this MTTF.

6. Obtain tabulated values for the failure rate for 10 different mission end times. The mission end times are 1,000 to 10,000 hours, using increments of 1,000 hours.

Solution

The following figure shows the data as entered in Weibull++, as well as the calculated parameters.

The following figures show the probability plot with the 90% two-sided confidence bounds and the pdf plot.

Both the reliability and MTTF can be easily obtained from the QCP. The QCP, with results, for both cases is shown in the next two figures.

To obtain tabulated values for the failure rate, use the Analysis Workbook or General Spreadsheet features that are included in Weibull++. (For more information on these features, please refer to the Weibull++ User's Guide. For a step-by-step example on creating Weibull++ reports, please see the Quick Start Guide). The following worksheet shows the mission times and the corresponding failure rates.

Suspension Data Example

19 units are being reliability tested and the following is a table of their times-to-failure and suspensions.

Non-Grouped Data Times-to-Failure Data with Suspensions
Data point index	Last Inspected	State End Time
1	F	2
2	S	3
3	F	5
4	S	7
5	F	11
6	S	13
7	S	17
8	S	19
9	F	23
10	F	29
11	S	31
12	F	37
13	S	41
14	F	43
15	S	47
16	S	53
17	F	59
18	S	61
19	S	67

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.07 \\ & {{{\hat{\sigma }}}_{T}}= & 28.41. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 46.40 \\ & {{{\hat{\sigma }}}_{T}}= & 28.64. \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 47.34 \\ & {{{\hat{\sigma }}}_{T}}= & 29.96. \end{align}\,\! }[/math]

Interval Censored Data Example

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Interval Data
Data point index	Last Inspected	State End Time
1	30	32
2	32	35
3	35	37
4	37	40
5	42	42
6	45	45
7	50	50
8	55	55

This is a sequence of interval times-to-failure data. Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 7.740. \end{align}\,\! }[/math]

For rank regression on x:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.40 \\ & {{{\hat{\sigma }}}_{T}}= & 9.03. \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on y (RRY) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 41.39 \\ & {{{\hat{\sigma }}}_{T}}= & 9.25. \end{align}\,\! }[/math]

The following plot shows the results if the data were analyzed using the rank regression on X (RRX) method.

Mixed Data Types Example

Suppose our data set includes left and right censored, interval censored and complete data, as shown in the following table.

Grouped Data Times-to-Failure with Suspensions and Intervals (Interval, Left and Right Censored)
Data point index	Number in State	Last Inspection	State (S or F)	State End Time
1	1	10	F	10
2	1	20	S	20
3	2	0	F	30
4	2	40	F	40
5	1	50	F	50
6	1	60	S	60
7	1	70	F	70
8	2	20	F	80
9	1	10	F	85
10	1	100	F	100

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 48.11 \\ & {{{\hat{\sigma }}}_{T}}= & 26.42 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 49.99 \\ & {{{\hat{\sigma }}}_{T}}= & 30.17 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 51.61 \\ & {{{\hat{\sigma }}}_{T}}= & 33.07 \end{align}\,\! }[/math]

Comparison of Analysis Methods

8 units are being reliability tested, and the following is a table of their failure times:

Non-Grouped Times-to-Failure Data
Data point index	State F or S	State End Time
1	F	2
2	F	5
3	F	11
4	F	23
5	F	29
6	F	37
7	F	43
8	F	59

Using the normal distribution and the maximum likelihood (MLE) parameter estimation method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 18.57 \end{align}\,\! }[/math]

If we analyze the data set with the rank regression on x (RRX) method, the computed parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 21.64 \end{align}\,\! }[/math]

For the rank regression on y (RRY) method, the parameters are:

[math]\displaystyle{ \begin{align} & \widehat{\mu }= & 26.13 \\ & {{{\hat{\sigma }}}_{T}}= & 22.28. \end{align}\,\! }[/math]