A random variable X has the following probability distribution: X… - Vidyadip

MCQ

A random variable X has the following probability distribution:

X	1	2	3	4
P(X)	$\frac{1}{7}$	$\frac{2}{7}$	$\frac{3}{7}$	$\frac{1}{7}$

Then, the variance of this distribution is

A
$\frac{49}{40}$
✓
$\frac{40}{49}$
C
$\frac{20}{29}$
D
$\frac{29}{20}$

✓

Answer

Correct option: B.

$\frac{40}{49}$

(B)
$\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}}, \mathrm{P}\left(x_{\mathrm{i}}\right)$
$=1\left(\frac{1}{7}\right)+2\left(\frac{2}{7}\right)+3\left(\frac{3}{7}\right)+4\left(\frac{1}{7}\right)=\frac{18}{7}$
$\begin{aligned} E\left(X^2\right)=\left(1^2\right)\left(\frac{1}{7}\right)+\left(2^2\right)\left(\frac{2}{7}\right)+\left(3^2\right)\left(\frac{3}{7}\right) +\left(4^2\right)\left(\frac{1}{7}\right)\end{aligned}$
$=\frac{1}{7}+\frac{8}{7}+\frac{27}{7}+\frac{16}{7}=\frac{52}{7}$
$\therefore \quad \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2$
$=\frac{52}{7}-\left(\frac{18}{7}\right)^2=\frac{40}{49}$

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