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)

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

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

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

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

$=10$