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{\bf Question}

Let $X$ and $Y$ denote the scores of two class tests for a
randomly selected student, called Miss T. Assume that $X$ and $Y$
is bivariate normal

$N_2(\mu_x=85,\mu_y=90,\sigma_x=10,\sigma_y=16,\rho=0.8)$.

\begin{description}
\item[(a)]
What is the probability that the sum of her score on the first two
tests will be greater than 200?

\item[(b)]
What is the probability that her score on the first test $(X)$
will be higher than her score on the second test?

\item[(c)]
If Miss T's score $X$ on the first test is 80, what is the
probability that her score on the second test will be higher than
90?

\end{description}


\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
Let $W=X+Y$

\begin{eqnarray*} E(W) & = & E(X) + E(Y)\\ & = &
\mu_x+\mu_y=85+90=175 \end{eqnarray*}

\begin{eqnarray*} {\rm var}(W) & = & {\rm var}(X)+{\rm var}(Y)+2 {\rm cov}(X,Y)\\ & = &
10^2+16^2+2(0.8)(10)(16)\\ & = & 612 \end{eqnarray*}

Since $W$ is a linear combination of $X$ and $Y$

$$W \sim N(175,612)$$

Therefore \begin{eqnarray*} P(W>200) & = & P
\left\{\frac{W-175}{\sqrt{612}}>\frac{200-175}{\sqrt{612}}\right\}\\
& = & 1-\Phi
\left(\frac{25}{\sqrt{612}}\right)=0.1562\end{eqnarray*}

\item[(b)]
Let $W=X-Y$

Therefore $E(W) = 85-90=-5$

${\rm var}(W)=10^2+16^2-2(0.8)(10)(16)=100$

$P(W>0) = P \left\{\frac{W-(-5)}{10}>\frac{0-(-5)}{10}\right\} =
1-\Phi (\frac{1}{2})=0.3085$

\item[(c)]
$P(Y>90|X=80)$

We need the distribution of $Y|X=x$.

$E(Y|X=x)=\mu_y+\rho\frac{\sigma_y}{\sigma_x}(x-\mu_x)$

${\rm var}(Y|X=x)=\sigma^2_y(1-\rho^2) = 16^2(1-0.8^2) = 92.16$

Therefore $E(Y|X=x)=90+0.8\frac{16}{10}(80-85) = 83.6$

Therefore \begin{eqnarray*} P(Y>90|X=80) & = &
P\left\{\left.\frac{Y-83.6}{9.6}>\frac{90-83.6}{9.6}\right|X=80\right\}\\
& = & P\left(Z>\frac{6.4}{9.6}\right)\ {\rm where}\ Z \sim
N(0,1)\\ & = & 0.2525 \end{eqnarray*}
\end{description}


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