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QUESTION

\begin{description}

\item[(b)]
The life (in months) of an industrial product is a continuous
random variable with probability density function

$$f_X(x)=\left\{\begin{array}{cl}
192(x+4)^{-4},&x>0,\\0,&\textrm{otherwise}.
\end{array}\right.$$

\begin{description}

\item[(i)]
Show that the distribution function is $\ds
F_X(x)=1-\frac{64}{(x+4)^3}$ for $x>0$.

\item[(ii)]
Find the probability that the life of the product is more than 4
months.

\item[(iii)]
Find the mean life of the product.

\end{description}

\end{description}

\bigskip

ANSWER

\begin{description}

\item[(b)]

$$f_X(x)=\left\{\begin{array}{cl}\frac{192}{(x+4)^4}&x>0\\0&\textrm{
otherwise}\end{array}\right.$$

\begin{description}

\item[(i)]
Distribution function
\begin{eqnarray*}
F_X(x)&=&\int_{-\infty}^x f_X(x)\,dx\\
&=&\int_{-\infty}^00\,dx+\int_0^x\frac{192}{(x+4)^4}\,dx\\
&=&192\left[\frac{(x+4)^{-3}}{-3}\right]_0^x\\
&=&64\left\{-\frac{1}{(x+4)^3}-(-\frac{1}{4^3})\right\}\\
F_X(x)&=&1-\frac{64}{(x+4)^3},\ x>0
\end{eqnarray*}

\item[(ii)]
$\ds
P(X>4)=1-P(X\leq4)=1-F_X(4)=1-\left(1-\frac{64}{(8)^3}\right)=\frac{64}{8^3}=\frac{1}{8}$

\item[(iii)]
\begin{eqnarray*}
{\rm Mean }&=&\int_{-\infty}^\infty xf_X(x)\,dx\\ &=&\int_0^\infty
x\frac{192}{(x+4)^4}\,dx\end{eqnarray*}

$\ds\int_0^X\frac{192x}{(x+4)^4}dx
=192\int_0^X\frac{x+4-4}{(x+4)^4}\,dx$

\begin{eqnarray*}
&=&192\int_0^X\left(\frac{1}{(x+4)^3}-\frac{4}{(x+4)^4}\right)\,dx\\
&=&192\left[\frac{(x+4)^{-2}}{-2}-\frac{4(x+4)^{-3}}{-3}\right]_0^X\\
&=&192\left(\frac{-1}{2(X+4)^2}+\frac{4}{3(X+4)^3}
+\frac{1}{2(4^2)}-\frac{4}{3(4^3)}\right)\end{eqnarray*}

Hence as $X\to\infty$
mean$=0+192\left(\frac{1}{32}-\frac{1}{48}\right)=2$.

So the mean is 2 weeks.

\end{description}

\end{description}



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