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QUESTION

\begin{description}

\item[(a)]
The probability density function of the random variable $X$ is
given by $$ f_X(x)=\left\{\begin{array}{ll}e^{-2cx}\textrm{ for
}x\geq0,\\0&\textrm{ otherwise. }\end{array}\right. $$

\begin{description}

\item[(i)]
Find the value of $c$.

\item[(ii)]
Find the mean of $X$.

\item[(iii)]
Show that the distribution function is $F_X(x)=1-e^{-x}$.

\end{description}

\item[(b)]
A company produces a video tape which lasts, on average, for 150
hours with a standard deviation of 36 hours. Assuming that the
life of a tape is normally distributed, find the probability that
any given tape will last less than 96 hours.

\end{description}


ANSWER

\begin{description}

\item[(a)]

\begin{description}

\item[(i)]
$\ds\int_{-\infty}^\infty f_X(x)\,dx=1 $ therefore
\begin{eqnarray*}
\int_{-\infty}^00\,dx+\int_0^\infty
e^{-2cx}\,dx&=&\left[\frac{e^{-2cx}}{-2c}\right]_0^\infty\\&=&
0-\left(\frac{1}{-2c}\right)=\frac{1}{2c}=1,\\ c&=&\frac{1}{2}
\end{eqnarray*}

\item[(ii)]
\begin{eqnarray*}{\rm Mean\ of\ }X &=&\mu_X= \int_{-\infty}^\infty
xf_X(x)\,dx\\&=& \int_0^xxe^{-x}\,dx\\&=&
\left[x\frac{e^{-x}}{-1}\right]_0^\infty-\int_0^\infty\left(-e^{-x}\right).1\,dx\\&=&
0-0+\left[\frac{e^{-x}}{-1}\right]_0^\infty\\&=& 0-(0-1)=1
\end{eqnarray*}

\item[(iii)]
\begin{eqnarray*}
F_X(x)&=&\int_{-\infty}^xf_X(x)\,dx=\int_0^x
e^{-x}\,dx\\&=&\left[\frac{e^{-x}}{-1}\right]_0^x=-e^{-x}-(-1)\\&=&1-e^{-x}
\end{eqnarray*}

\end{description}

\item[(b)]
We are given that $\mu_X=150,\ \sigma_X=36$. Put $\ds
Z=\frac{X-\mu}{36}$ then $Z\sim N(0,1)$

\begin{eqnarray*}
P(X<96)&=&P\left(Z<\frac{96-150}{36}\right)\\ &=&P(Z<-1.5)
\\&=&P(Z>1.5)\textrm{, by symmetry}\\
&=&1-P(Z\leq1.5)\\ &=&1-0.9332=0.0668
\end{eqnarray*}

\end{description}



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