By the proposition of the mean and variance of a random variable $X$ having gamma distribution $f(x;\alpha,\beta)$,
\[
E(X)=\mu=\alpha\beta,\qquad V(X)=\sigma^2=\alpha\beta^2
\]
Hence, from the given condition, $\mu=24$ and $\sigma=12$. Then, we have
\begin{eqnarray}
\alpha\beta&=&\mu=24,\\
\alpha\beta^2&=&\sigma^2=12^2
\end{eqnarray}
It is easy to get the solution $\alpha=4$ and $\beta=6$.
(a)
The probability that a transistor will last between $12$ and $24$ weeks is
\[
\begin{split}
P(12\leqslant X\leqslant 24)&=F(24;\alpha,\beta)-F(12;\alpha,\beta)\\
&=F(\frac{24}{\beta};\alpha)-F(\frac{12}{\beta};\alpha)\\
&=F(\frac{24}{6};4)-F(\frac{12}{6};4)\\
&=F(4;4)-F(2;4)\\
&=.567-.143\\
&=.424
\end{split}
\]
(b)
The probability that a transistor will last at most $24$ weeks is
\[
P(X\leqslant 24)=F(24;\alpha,\beta)=F(\frac{24}{\beta};\alpha)=F(\frac{24}{6};4)=F(4;4)=.567
\]
Let $m$ be the median of the lifetime distribution, then it satisfies
\[
P(X\leqslant m)=0.5.
\]
Since $P(X\leqslant 24)=.567>.5$, we get $m < 24$.
So, the answer of the question is yes.
(c)
Let $P(X\leqslant x)=0.99$, then
\[
0.99=P(X\leqslant x)=F(x;\alpha,\beta)=F(\frac{x}{\beta};\alpha)=F(\frac{x}{6};4)
\]
By Table A.4, we have $F(10;4)=0.99$, then $\frac{x}{6}=10$, it infers that $x=60$.
Therefore, the 99th percentile of the lifetime distribution is $60$.
(d)
By condition, $P(X\geqslant t)=.5\%$, then
\[
\begin{split}
&1-P(X < t)=0.005\\
\Rightarrow&1-P(X\leqslant t)=0.005\\
\Rightarrow&P(X\leqslant t)=0.995\\
\Rightarrow&F(t;\alpha,\beta)=0.995\\
\Rightarrow&F(\frac{t}{\beta};\alpha)=0.995\\
\Rightarrow&F(\frac{t}{6};4)=0.995
\end{split}
\]
Since $F(11;4)=0.995$ by Table A.4, we get $\frac{t}{6}=11$. Therefore $t=66$.
