Successive weekly sales, in units of $1,000$, have a bivariate normal distribution with common mean $40$, common standard deviation $6$, and correlation .$6$.

(a) Find the probability that the total of the next $2$ weeks’ sales exceeds $90$.

(b) If the correlation were .$2$rather than .$6$, do you think that this would increase or decrease the answer to (a)? Explain your reasoning.

(c) Repeat (a) when the correlation is $2$.

Random variables $X$and $Y$that marks the sales in two successive weeks.

$E(X+Y)=E\left(X\right)+E\left(Y\right)=80$

and $Var(X+Y)=Var\left(X\right)+Var\left(Y\right)+2Cov(X,Y)$

$=115.2$

$=36*06$

$=21.6$

So $X+Y$ is Normally distributed random variable with parameters

We are required to find:

$P(X+Y>90)=P\left(\frac{X+Y-80}{\sqrt{115.2}}>\frac{90-80}{\sqrt{115.2}}\right)$

$=1-P\left(\frac{X+Y-80}{\sqrt{115.2}}\le \frac{90-80}{\sqrt{115.2}}\right)$

$=1-\Phi (0.931675)$

$=0.1757$

The probability that the total of the next $2$weeks’ sales exceeds $90$is$0.1757$.

Random variables $X$and $Y$that marks the sales in two successive weeks.

$E(X+Y)=E\left(X\right)+E\left(Y\right)=80$.

The probability would decrease and the reason is following. Observe that $E(X+Y)=80<90\mathrm{s}$ these variables have to be greater than their mean if they want to achieve $90$ or more. If the correlation is lower, the rising of $X$ would not imply so big rising of $Y$ as it would be with greater correlation: Therefore, it's much harder to obtain $90$ or more with lower correlation than with bigger correlation, so the probability must decrease.

The probability must be decreasing.

Random variables$X$and$Y$that marks the sales in two successive weeks$E(X+Y)=80$.

Recalculate the variance,

$Var(X+Y)=Var\left(X\right)+Var\left(Y\right)+2Cov(X,Y)$

$=86.4$

$Cov(X,Y)=\sqrt{Var\left(X\right)}\sqrt{Var\left(Y\right)}\rho (X,Y)$

$=7.2$

So, $X+Y$is Normally distributed random variable with parameters,

$X+Y~\mathcal{N}(80,86.4)$.

$P(X+Y>90)=P\left(\frac{X+Y-80}{\sqrt{86.4}}>\frac{90-80}{\sqrt{86.4}}\right)$

$=1-P\left(\frac{X+Y-80}{\sqrt{86.4}}\le \frac{90-80}{\sqrt{86.4}}\right)$

$=1-\Phi (1.0758287)$

$=0.141$

The correlation is$0.141$.