Question

For the following probability distribution:

X	$1$	$2$	$3$	$4$
P(X)	$\frac{1}{10}$	$\frac{1}{5}$	$\frac{3}{10}$	$\frac{2}{5}$

$E(X^2)$ is equal to

• A. $3$
• B. $5$
• C. $7$
• D. $10$

Answer 1

Answer: (D)

$\Sigma\left(X^{2}\right) = \Sigma X^{2}P\left(X\right)$

$= \left(1\right)^{2}\times\frac{1}{10} +\left(2\right)^{2}\times\frac{1}{5} +\left(3\right)^{2}\times\frac{3}{10} +\left(4\right)^{2}\times\frac{2}{5}$

$=\frac{1}{10} +\frac{4}{5} +\frac{27}{10} +\frac{32}{5}$

$ =\frac{1+8+27+64 }{10} = \frac{100}{10}$

$ = 10$