Question

Suppose that X is a random variable with mean 20 and standard deviation 5. Also

Answer 1

Disregard my earlier question.

Recall that the variance of a random variable $X$, denoted $\mathbb V(X)$, is given by

$\mathbb E((X-\mu)^2)$

where $\mathbb E(X)$ denotes the expected value/mean of $X$, and $\mu=\mathbb E(X)$ is the actual mean of $X$.

Now, recall that

$\mathbb V(X)=\mathbb E((X-\mu)^2)=\mathbb E(X^2-2\mu X+\mu^2)$
$\mathbb V(X)=\mathbb E(X^2)-2\mu^2+\mu^2$
$\mathbb V(X)=\mathbb E(X^2)-\mathbb E(X)^2$

(a) $Z=2+10X$

$\mathbb V(Z)=\mathbb E((2+10X)^2)-\mathbb E(2+10X)^2$
$\mathbb V(Z)=\mathbb E(4+40X+100X^2)-(\mathbb E(2)+\mathbb E(10X))^2$
$\mathbb V(Z)=\mathbb E(4)+40\mathbb E(X)+100\mathbb E(X^2)-(\mathbb E(2)+10\mathbb E(X))^2$
$\mathbb V(Z)=100\mathbb E(X^2)-40000$
$\mathbb V(Z)=100\left(\mathbb E(X^2)-\mathbb E(X)^2+\mathbb E(X)^2\right)-40000$
$\mathbb V(Z)=100\left(\mathbb V(X)+\mathbb E(X)^2\right)-40000$
$\mathbb V(Z)=100\left(5^2+400\right)-40000$
$\mathbb V(Z)=2500$

The standard deviation is the square root of the variance, so for (a) you have

$\sqrt{\mathbb V(Z)}=\sqrt{2500}=50$

That should give you an idea as to how to figure out the rest.