Finding E((X-2)²)

Question

This is a practice quiz as I have a test in a few days. I've already completed the first and second questions and my answers are: (1) h = 0.4 (2) k = 30

I'm struggling on question 3, I know how to find E(X) using the x•p(x) method but how do I find E((X-2)²). I tried (X-2)²•p(x) but I'm not sure that it's the correct answer. Any help is greatly appreciated!]1

Answer 1

$\begin{align}E[(X-2)^2] &= E[X^2-4X+4]\\ &= E[X^2]-4E[X]+E[4]\\ &= \sum_xx^2p(x)-4\sum_xxp(x)+4\\ &= 93.8 - 4\times4+4\\ &= 81.8 \end{align}$

On a side note: $E[g(X)]=\displaystyle{\sum_x}g(x)p(x)$.

Answer 2

(a) We require the sum of all probabilities to be $1$, so $h = 0.4$.

(b) We require $$4 = \operatorname{E}[X] = (-1)(0.2) + 0(0.3) + k(0.1) + (3)(0.4) = 1 + k/10,$$ hence $k = 30$.

(c) We compute $$\operatorname{E}[(X-2)^2] = (-1-2)^2(0.2) + (0-2)^2(0.3) + (30-2)^2 (0.1) + (3-2)^2 (0.4) = 81.8.$$

It is worth noting that the value of $80.6$ is obtained if you do not include the second term, $(0-2)^2 (0.3)$. It seems likely that whoever computed the solution failed to realize that even if $X = 0$ contributes nothing to the expectation, once you compute $(X-2)^2 = 4$, this does contribute. They probably dropped the term because it was dropped when computing $\operatorname{E}[X]$.