Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section A

Chapter

Probability

Question:

Let X be a random variable whose probability distribution is given by the table

X 1 3 5 7
P(X) $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$

Then variance of $x$ is

Options:

$\sqrt{\frac{57}{3}}$

$\frac{19}{3}$

$4$

$\frac{67}{3}$

Correct Answer:

$\frac{19}{3}$

Explanation:
X 1 3 5 7
$X^2$ 1 9 25 49
P(X) $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$

mean = $E(X) = ∑pixi = 1×\frac{1}{3}+3×\frac{1}{6}+5×\frac{1}{6}+7×\frac{1}{3}$

$E(X) =\frac{2+3+5+7}{6}=\frac{24}{6}=4=E(X)$

so $E(X^2) =∑pi{x_i}^2 =1×\frac{1}{3}+9×\frac{1}{6}+25×\frac{1}{6}+49×\frac{1}{3}$

$E(X^2) =\frac{2+9+25+98}{6}=\frac{134}{6}$

$σ^2$ (variance) = $E(X^2)-(E(X))^2$

$=\frac{134}{6}-4^2=\frac{67}{3}-16$

$=\frac{67-48}{3}=\frac{19}{3}$