Question

Let X be a random variable with the following probability distribution.
x -3 6 9
f(x) 1/6 1/2 1/3
(a) Find μg(x), where g(X)=(2X+1)2.
(b) Find the standard deviation of g(X)

Answer 1

It looks like you're saying

$f_X(x) = \begin{cases}\dfrac16&\text{if }x=-3 \\\\ \dfrac12&\text{if }x=6\\\\\dfrac13&\text{if }x=9\\\\0&\text{otherwise}\end{cases}$

(a) Given g(X) = (2X + 1)², the mean of g(X) is

$E[g(X)] = \displaystyle \sum_{x\in\{-3,6,9\}} g(x) f_X(x) \\\\ = g(-3)f_X(-3) + g(6)f_X(6)+g(9)f_X(9) = \frac{25}6 + \frac{169}2 + \frac{361}3 = \boxed{209}$

(b) First compute the second moment of g(X) :

$E[g(X)^2] = \displaystyle \sum_{x\in\{-3,6,9\}} g(x)^2 f_X(x) \\\\ = g(-3)^2f_X(-3) + g(6)^2f_X(6)+g(9)^2f_X(9) = \frac{625}{6} + \frac{28561}{2} + \frac{130321}{3} = 57825$

Then the variance of g(X) is

$\mathrm{Var}[g(X)] = E[g(X)^2] - E[g(X)]^2 = 57825 - 209^2 = 14144$

The standard deviation is the square root of the variance:

$\sqrt{\mathrm{Var}[g(X)]} = \sqrt{14144} \approx \boxed{118.93}$