Suppose the time (in hours) taken by a homeowner to mow his lawn(修剪他的草坪) in an rv $X$ having a gamma distribution with parameters $\alpha=2$ and $\beta=\frac{1}{2}$. What is the probability that it takes:
1
The pdf is
\[
f(x;\alpha,\beta)=\begin{cases}
\frac{1}{\beta^{\alpha}\Gamma(\alpha)}x^{\alpha-1}\cdot e^{-\frac{x}{\beta}}, & x\geqslant 0,\\
0, &\text{otherwise}.
\end{cases}
\]
Here $\alpha=2$ and $\beta=\frac{1}{2}$.
(a) (方法一)
\[
\begin{split}
P(X\leqslant 1)&=\int_{0}^{1}f(x;\alpha,\beta)dx=\int_{0}^{1}\frac{1}{(\frac{1}{2})^2\Gamma(2)}x^{2-1}e^{-\frac{x}{\frac{1}{2}}}dx\\
&=\int_{0}^{1}\frac{4}{\Gamma(2)}xe^{-2x}dx=\frac{4}{\Gamma(2)}\int_{0}^{1}xe^{-2x}dx\\
&=\frac{4}{1}\int_{0}^{1}\frac{1}{-2}xde^{-2x}=(-2)\cdot\int_{0}^{1}xde^{-2x}\\
&=(-2)\cdot\biggl[xe^{-2x}\biggr|_{0}^{1}-\int_{0}^{1}e^{-2x}dx\biggr]\\
&=(-2)\cdot\biggl[e^{-2}-\frac{1}{-2}e^{-2x}\biggr|_{0}^{1}\biggr]\\
&=-2\biggl[e^{-2}+\frac{1}{2}(e^{-2}-1)\biggr]=-2\Bigl[\frac{3}{2}e^{-2}-\frac{1}{2}\Bigr]\\
&=1-3e^{-2}\\
&\approx 0.59399415
\end{split}
\]
(方法二)
By the proposition, the cdf of the rv $X$ which having a gamma distribution is
\[
F(x;\alpha,\beta)=F(\frac{x}{\beta};\alpha).
\]
Then, we have
\[
P(X\leqslant 1)=F(1;2,\frac{1}{2})=F(\frac{1}{\frac{1}{2}};2)=F(2;2)\approx 0.594
\]
For the value of $F(2;2)$, we can search the table in the Appdix.
(b)
The probability that it takes at least $2$ hours to mow the lawn is
\[
\begin{split}
P(X\geqslant 2)&=1-P(X < 2)=1-P(X\leqslant 2)\\
&=1-F(2;2,\frac{1}{2})=1-F(\frac{2}{\frac{1}{2}};2)\\
&=1-F(4;2)=1-0.908\\
&=0.092
\end{split}
\]
(c)
The probability that it takes between $.5$ and $1.5$ hours to mow the lawn is
\[
\begin{split}
P(.5\leqslant X\leqslant 1.5)&=F(1.5;\alpha,\beta)-F(.5;\alpha,\beta)\\
&=F(\frac{1.5}{\beta};\alpha)-F(\frac{.5}{\beta};\alpha)\\
&=F(\frac{1.5}{\frac{1}{2}};2)-F(\frac{.5}{\frac{1}{2}};2)\\
&=F(3;2)-F(1;2)\\
&=.801-.264\\
&=0.537
\end{split}
\]