The Lognormal Distribution

=The Lognormal Distribution=

The lognormal distribution is commonly used to model the lives of units whose failure modes are of a fatigue-stress nature. Since this includes most, if not all, mechanical systems, the lognormal distribution can have widespread application. Consequently, the lognormal distribution is a good companion to the Weibull distribution when attempting to model these types of units. As may be surmised by the name, the lognormal distribution has certain similarities to the normal distribution. A random variable is lognormally distributed if the logarithm of the random variable is normally distributed. Because of this, there are many mathematical similarities between the two distributions. For example, the mathematical reasoning for the construction of the probability plotting scales and the bias of parameter estimators is very similar for these two distributions.

Example 9
Determine the lognormal parameter estimates for the data given in Table 9.3.

Solution to Example 9
This is a sequence of interval times-to-failure where the intervals vary substantially in length. Using Weibull++, the computed parameters for maximum likelihood are calculated to be:


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 3.64 \\ & {{{\hat{\sigma }}}_}= & 0.18 \end{align}$$

For rank regression on $$X\ \ :$$


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 3.64 \\ & {{{\hat{\sigma }}}_}= & 0.17 \end{align}$$

For rank regression on $$Y\ \ :$$


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 3.64 \\ & {{{\hat{\sigma }}}_}= & 0.21 \end{align}$$

Example 10
Determine the lognormal parameter estimates for the data given in Table 9.4.

Solution to Example 10
Using Weibull++, the computed parameters for maximum likelihood are:


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 2.83 \\ & {{{\hat{\sigma }}}_}= & 1.10 \end{align}$$

For rank regression on $$X\ \ :$$


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 2.83 \\ & {{{\hat{\sigma }}}_}= & 1.24 \end{align}$$

For rank regression on $$Y\ \ :$$


 * $$\begin{align}

& {{{\hat{\mu }}}^{\prime }}= & 2.83 \\ & {{{\hat{\sigma }}}_}= & 1.36 \end{align}$$

Example 11
From Kececioglu [19, p. 406]. Nine identical units are tested continuously to failure and their times-to-failure were recorded at 30.4, 36.7, 53.3, 58.5, 74.0, 99.3, 114.3, 140.1, and 257.9 hours.

Solution to Example 11
The results published were obtained by using the unbiased model. Published Results (using MLE):


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=4.3553 \\ {{\widehat{\sigma }}_}=0.67677 \\ \end{matrix}$$

This same data set can be entered into Weibull++ by creating a data sheet capable of handling non-grouped time-to-failure data. Since the results shown above are unbiased, the Use Unbiased Std on Normal Data option in the User Setup must be selected in order to duplicate these results. Weibull++ computed parameters for maximum likelihood are:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=4.3553 \\ {{\widehat{\sigma }}_}=0.6768 \\ \end{matrix}$$

Example 12
From Kececioglu [20, p. 347]. Fifteen identical units were tested to failure and following is a table of their times-to-failure:

$$\text{Table 9}\text{.5 - Data of Example 11}$$

$$\begin{matrix} \text{Data Point Index} & \text{Time-to-Failure, hr} \\ \text{1} & \text{62}\text{.5} \\ \text{2} & \text{91}\text{.9} \\ \text{3} & \text{100}\text{.3} \\ \text{4} & \text{117}\text{.4} \\ \text{5} & \text{141}\text{.1} \\ \text{6} & \text{146}\text{.8} \\ \text{7} & \text{172}\text{.7} \\ \text{8} & \text{192}\text{.5} \\ \text{9} & \text{201}\text{.6} \\ \text{10} & \text{235}\text{.8} \\ \text{11} & \text{249}\text{.2} \\ \text{12} & \text{297}\text{.5} \\ \text{13} & \text{318}\text{.3} \\ \text{14} & \text{410}\text{.6} \\ \text{15} & \text{550}\text{.5} \\ \end{matrix}$$

Solution to Example 12
Published results (using probability plotting):


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=5.22575 \\ {{\widehat{\sigma }}_}=0.62048. \\ \end{matrix}$$

Weibull++ computed parameters for rank regression on X are:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=5.2303 \\ {{\widehat{\sigma }}_}=0.6283. \\ \end{matrix}$$

The small differences are due to the precision errors when fitting a line manually, whereas in Weibull++ the line was fitted mathematically.

Example 13
From Nelson [30, p. 324]. Ninety-six locomotive controls were tested, 37 failed and 59 were suspended after running for 135,000 miles. Table 9.6 (at the end of this chapter) shows their times-to-failure.

Solution to Example 13
The distribution used in the publication was the base-10 lognormal. Published results (using MLE):


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=2.2223 \\ {{\widehat{\sigma }}_}=0.3064 \\ \end{matrix}$$

Published 95% confidence limits on the parameters:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=\left\{ 2.1336,2.3109 \right\} \\ {{\widehat{\sigma }}_}=\left\{ 0.2365,0.3970 \right\} \\ \end{matrix}$$

Published variance/covariance matrix:


 * $$\left[ \begin{matrix}

\widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0020 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma }}}_})=0.001 \\ {} & {} & {} \\   \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma }}}_})=0.001 & {} & \widehat{Var}\left( {{{\hat{\sigma }}}_} \right)=0.0016  \\ \end{matrix} \right]$$

To replicate the published results (since Weibull++ uses a lognormal to the base $$e$$ ), take the base-10 logarithm of the data and estimate the parameters using the Normal distribution and MLE.

•	Weibull++ computed parameters for maximum likelihood are:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=2.2223 \\ {{\widehat{\sigma }}_}=0.3064 \\ \end{matrix}$$

•	Weibull++ computed 95% confidence limits on the parameters:


 * $$\begin{matrix}

{{\widehat{\mu }}^{\prime }}=\left\{ 2.1364,2.3081 \right\} \\ {{\widehat{\sigma }}_}=\left\{ 0.2395,0.3920 \right\} \\ \end{matrix}$$

•	Weibull++ computed/variance covariance matrix:


 * $$\left[ \begin{matrix}

\widehat{Var}\left( {{{\hat{\mu }}}^{\prime }} \right)=0.0019 & {} & \widehat{Cov}({{{\hat{\mu }}}^{\prime }},{{{\hat{\sigma }}}_})=0.0009 \\ {} & {} & {} \\   \widehat{Cov}({\mu }',{{{\hat{\sigma }}}_})=0.0009 & {} & \widehat{Var}\left( {{{\hat{\sigma }}}_} \right)=0.0015  \\ \end{matrix} \right]$$