Question

If $E\left[X\right]=1$and $Var\left(X\right)=5$find

(a)$E\left[\right(2+X^2\left)\right]$

(b)$Var(4+3X)$

Short answer

a) The value of $E\left[\left(2+X^2\right)\right]$is $14$.

b) The value of$Var(4+3X)$is$45.$

Step-by-step solution

Given information (Part a)

$E\left(X\right)=1$

$Var\left(X\right)=5$

Then$E\left[\left(2+X^2\right)\right]=?$

Explanation (Part a) 

We know that

$\mathrm{E}\left({\mathrm{X}}^2\right)=Var\left(\mathrm{X}\right)+\left(\mathrm{E}\right(\mathrm{X}){)}^2$

$=5+1$

$=6$

Calculate the expected value of $E\left[\left(2+X^2\right)\right]$

$\mathrm{E}\left[(2+\mathrm{X}{)}^2\right]=\mathrm{E}\left[4+{\mathrm{X}}^2+4\mathrm{X}\right]$

$=4+6+4$

$=14$

Final Answer (Part a) 

The value of $E\left[\left(2+X^2\right)\right]$is$14.$

Given information (Part b)

$E\left(X\right)=1$

$Var\left(X\right)=5$

Then$Var(4+3x)=?$

Explanation (Part b)  

We know that,

$E\left(X^2\right)=Var\left(X\right)+\left(E\right(X){)}^2$

$=5+1$

role="math" localid="1647519062218" $=6$

Calculate the variance.

$Var(4+3X)=0+9Var\left(X\right)$

$=(9\times 5)$

$=45$

Final Answer (Part b)  

Hence, the value of$Var(4+3x)$is$45.$