Answer to Question #157975 in Statistics and Probability for Lusy

Question

Let X be a random variable with the following probability distribution:

x : 3 6 9

P(X=x) : 1/6 1/2 1/3

Find E(X) and E(X^2)and using the laws of expectation, evaluate E . (2X+1)^2.

Answer 1

$E(X)=\sum{_{i=3}^9}XiP(Xi)$

$=3(\frac{1}{6})+6(\frac{1}{2})+9(\frac{1}{3})$

$=\frac{1}{2}+3+3$

$=6\frac{1}{2}$

$E(X^2)=\sum{_{i=3}^9}Xi^2P(Xi)$

$=9(\frac{1}{6})+36(\frac{1}{2})+81(\frac{1}{3})$

$=\frac{3}{2}+18+27$

$=46\frac{1}{2}$

$E(2X+1)^2=E[(2X+1)(2X+1)]$

$=E(4X^2+4X+1)$

$=4E(X^2)+4E(X)+1$

$=(46\frac{1}{2}×4)+(4×6\frac{1}{2})+1$

$=186+26+1$

$=213$