Template:Alta weibull distribution

The Weibull Distribution
The Weibull distribution is one of the most commonly used distributions in reliability engineering because of the many shapes it attains for various values of $$\beta $$  (slope). It can therefore model a great variety of data and life characteristics [18]. The 2-parameter Weibull $$pdf$$  is given by:
 * $$f(T)=\frac{\beta }{\eta }{{\left( \frac{T}{\eta } \right)}^{\beta -1}}{{e}^{-{{\left( \tfrac{T}{\eta } \right)}^{\beta }}}}$$


 * where:


 * $$f(T)\ge 0,\text{ }T\ge 0,\text{ }\beta >0,\text{ }\eta >0\text{ }$$


 * and:
 * •	 $$\eta =$$ scale parameter.
 * •	 $$\beta =$$ shape parameter (or slope).

Parameter Estimation
The estimates of the parameters of the Weibull distribution can be found graphically on probability plotting paper, or analytically using either least squares or maximum likelihood. (Parameter estimation methods are presented in detail in Appendix B.)

Example 3
Let's assume six identical units are being reliability tested at the same application and operation stress levels. All of these units fail during the test after operating the following times (in hours), $${{T}_{i}}$$ : 93, 34, 16, 120, 53 and 75. The steps for determining the parameters of the Weibull $$pdf$$  representing the data, using probability plotting, are as follows:
 * •	Rank the times-to-failure in ascending order as shown next.

$$\begin{matrix} \text{Time-to-} & \text{Failure Order Number} \\ \text{failure, hrs} & \text{out of a Sample Size of 6} \\ \text{16} & \text{1} \\ \text{34} & \text{2} \\ \text{53} & \text{3} \\ \text{75} & \text{4} \\ \text{93} & \text{5} \\ \text{120} & \text{6} \\ \end{matrix}$$
 * •	Obtain their median rank plotting positions. The times-to-failure, with their corresponding median ranks, are shown next.

$$\begin{matrix} \text{Time-to-} & \text{Median} \\ \text{failure, hr} & \text{Rank, }% \\ \text{16} & \text{10}\text{.91} \\ \text{34} & \text{26}\text{.44} \\ \text{53} & \text{42}\text{.14} \\ \text{75} & \text{57}\text{.86} \\ \text{93} & \text{73}\text{.56} \\ \text{120} & \text{89}\text{.10} \\ \end{matrix}$$
 * •	On a Weibull probability paper, plot the times and their corresponding ranks. Fig. 11 displays an example of a Weibull probability paper (the solution is given in Fig. 12).


 * •	Draw the best possible straight line through the plotted points (as shown in Fig. 12).
 * •	Obtain the slope of this line by drawing a line, parallel to the one just obtained, through the slope indicator. This value is the estimate of the shape parameter $$\widehat{\beta }$$ . In this case  $$\widehat{\beta }=1.4$$.
 * •	At the $$Q(t)=63.2%$$  ordinate point, draw a straight horizontal line until this line intersects the fitted straight line. Draw a vertical line through this intersection until it crosses the abscissa. The value at the intersection of the abscissa is the estimate of  $$\widehat{\eta }$$ . For this case  $$\widehat{\eta }=76$$  hr. (This is always at 63.2% since  $$Q(T)=1-{{e}^{-{{(\tfrac{\eta }{\eta })}^{\beta }}}}=1-{{e}^{-1}}=0.632=63.2%).$$

Now any reliability value for any mission time $$t$$  can be obtained. For example, the reliability for a mission of 15 hr, or any other time, can now be obtained either from the plot or analytically (i.e. using the equations given in Section 5.2.1).

To obtain the value from the plot, draw a vertical line from the abscissa, at $$t=15$$  hr, to the fitted line. Draw a horizontal line from this intersection to the ordinate and read $$Q(t)$$, in this case  $$Q(t=15)=9.8%$$. Thus, $$R(t=15)=1-Q(t)=90.2%$$. This can also be obtained analytically from the Weibull reliability function since both of the parameters are known.
 * $$R(t=15)={{e}^{-{{\left( \tfrac{15}{\eta } \right)}^{\beta }}}}={{e}^{-{{\left( \tfrac{15}{76} \right)}^{1.4}}}}=90.2%.$$