Question

a. compute the expected value and the variance for $x$ and $y$.
$e(x)=$
$e(y)=$
$var(x)=$
$var(y)=$
b. develop a probability distribution for $x + y$ (to 2 decimals).
$x + y$ $f(x + y)$
150 0.30
60 0.60
110 0.10
c. using the result of part (b), compute $e(x + y)$ and $var(x + y)$.
$e(x + y)=$
$var(x + y)=$
d. compute the covariance and correlation for $x$ and $y$. if required, round your answers to 1 decimal places.

Explanation:

Step1: Recall expected - value formula for discrete random variables

The expected value of a discrete random variable $Z$ is $E(Z)=\sum zf(z)$. For $E(x + y)$ using the probability - distribution of $x + y$ from part (b):
$E(x + y)=\sum_{i}(x + y)_if((x + y)_i)$
$=(60\times0.6)+(110\times0.1)+(150\times0.3)$
$=36 + 11+45$
$=92$

Step2: Recall variance formula for discrete random variables

The variance of a discrete random variable $Z$ is $Var(Z)=\sum(z - E(Z))^{2}f(z)$.
First, find $(x + y)-E(x + y)$ for each value of $x + y$:
For $x + y = 60$: $60 - 92=-32$; for $x + y = 110$: $110 - 92 = 18$; for $x + y = 150$: $150 - 92 = 58$
Then, $Var(x + y)=(60 - 92)^{2}\times0.6+(110 - 92)^{2}\times0.1+(150 - 92)^{2}\times0.3$
$=(-32)^{2}\times0.6+18^{2}\times0.1 + 58^{2}\times0.3$
$=(1024\times0.6)+(324\times0.1)+(3364\times0.3)$
$=614.4+32.4 + 1009.2$
$=1656$

Step3: Recall covariance formula

The covariance $Cov(x,y)=E(xy)-E(x)E(y)$. We can also use the formula $Var(x + y)=Var(x)+Var(y)+2Cov(x,y)$
We know $Var(x)=189$, $Var(y)=729$, and $Var(x + y)=1656$
$1656=189 + 729+2Cov(x,y)$
$2Cov(x,y)=1656-(189 + 729)$
$2Cov(x,y)=1656 - 918$
$2Cov(x,y)=738$
$Cov(x,y)=369$

Step4: Recall correlation formula

The correlation coefficient $
ho_{xy}=\frac{Cov(x,y)}{\sqrt{Var(x)Var(y)}}$
$
ho_{xy}=\frac{369}{\sqrt{189\times729}}$
$=\frac{369}{\sqrt{137781}}$
$=\frac{369}{371.2}$
$\approx0.99$

Answer:

$E(x + y)=92$
$Var(x + y)=1656$
$Cov(x,y)=369$
$
ho_{xy}\approx0.99$