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Probability is the likelihood that an event will occur and is calculated by dividing the number of favourable outcomes by the total number of possible outcomes. The simplest example is a coin flip. When you flip a coin there are only two possible outcomes, the result is either heads or tails.

It is assumed that a transistor's lifetime (in weeks) is designated by a number \(X\) and follows a gamma distribution.

\(\mu = 24\)

\(\sigma = 12\)

As we all know, the mean and standard deviation of a gamma distribution are determined by the parameters.

\({\bf{\alpha }}\)and \(\beta \) as follows :

\(\mu = \alpha \cdot \beta \)

\(\sigma = \beta \sqrt \alpha \)

As a result of the aforementioned two equations, we can calculate the values of \(\alpha \)and \(\beta \)

\(\alpha = \frac{{{\mu ^2}}}{{{\sigma ^2}}} = \frac{{24}}{{12}}\)

\(\alpha = 4\)

\(\beta = \frac{{{\sigma ^2}}}{\mu } = \frac{{{{12}^2}}}{{24}}\)

\(\beta = 6\)

(a)

\(P(12 < X < 24)\)is the likelihood that a transistor will last between 12 and 24 weeks.

\({\rm{P(12 < X < 24) = F(24 ; 4,6) - F(12 ; 4,6)}}\)

\( = F\left( {\frac{{24}}{6};4} \right) - F\left( {\frac{{12}}{6};4} \right)\)

\( = F(4;4) - F(2;4)\;\;\;{\rm{ (Use Appendix A - 4 here) }}\)

\({\rm{ = 0}}{\rm{.567 - 0}}{\rm{.143}}\)

\({\rm{P(12 < X < 24) = 0}}{\rm{.424}}\)

Proposition: Assume \(X\)has a gamma distribution with \(\alpha \) and \(\beta \) as parameters. The cdf of \(X\)is then provided by for any \(x > 0\).

\(P(X \le x) = F(x;\alpha ,\beta ) = F\left( {\frac{x}{\beta };\alpha } \right)\)

The incomplete gamma function is \(F(x;\alpha )\).

(b)

\(P(X \le 24)\)is the chance that a transistor will last at most 24 weeks.

\(P(X \le 24) = F(24;4,6)\)

\( = F\left( {\frac{{24}}{6};4} \right)\)

\( = F(4;4)\;\;\;{\rm{ (Use Appendix A - 4 here) }}\)

\(P(X \le 24) = 0.567\)

Since we know that the median \((\tilde \mu )\) of the distribution is less than 24.

(c)

If we use \({\eta _{0.99}}\) to represent the \({99^{{\rm{th }}}}\) percentile of the lifetime distribution, we may deduce that

\(P\left( {X \le {\eta _{0.99}}} \right) = 0.99\)

However, because the lifespan distribution is a gamma distribution with \(\alpha \) and \(\beta \) equal to 4 and 6, it is a gamma distribution.

Hence

\(P\left( {X \le {\eta _{0.99}}} \right) = F\left( {{\eta _{0.99}};4,6} \right) = F\left( {\frac{{{\eta _{0.99}}}}{6};4} \right)\)

We now discover the row corresponding to column \(\alpha = 4\) and value \(0.99\) in Appendix A-4, which turns out to be the \(x = 10\) row.

\(\frac{{{\eta _{0.99}}}}{6} = 10\)

\({\eta _{0.99}} = 60\)

Therefore, the 99th percentile of the lifetime distribution is \({\rm{60}}\)

(d)

Only \(0.5\% \)of all transistors are expected to be operational after \(t\) weeks. This suggests that \(99.5\% \)of them have a lifetime of less than that. As a result, \(t\) is the \({99.5^{{\rm{th }}}}\)percentile.

\(P(X \le t) = 0.995\)

Also

\(P(X \le t) = F(t;4,6) = F\left( {\frac{t}{6};4} \right)\)

Now in Appendix A-4, we find the row corresponding to column \(\alpha = 4\) and value \(0.995,\)which turn out to be \(x = 11\) row.

\(\frac{t}{6} = 11\)

\({\rm{t = 66}}\)

Therefore, the value of t is \({\rm{66}}\).