Let X be a random variable with distribution. x -2 -1 3 4 6 P(X=x) 1 5 a 1 3 1 5 ~b If the mean of X is 2.3 and variance of X is ^ 2 , then 100 ^ 2 is equal to :

Let X be a random variable with distribution. x -2 -1 3 4 6 P(X=x…

MCQ

Let $\mathrm{X}$ be a random variable with distribution.

$\mathrm{x}$	$-2$	$-1$	$3$	$4$	$6$
$\mathrm{P}(\mathrm{X}=\mathrm{x})$	$\frac{1}{5}$	$\mathrm{a}$	$\frac{1}{3}$	$\frac{1}{5}$	$\mathrm{~b}$

If the mean of $X$ is $2.3$ and variance of $X$ is $\sigma^{2}$, then $100 \sigma^{2}$ is equal to :

✓

Correct option: A.

$781$

$\mathrm{x}$	$-2$	$-1$	$3$	$4$	$6$
$\mathrm{P}(\mathrm{X}=\mathrm{x})$	$\frac{1}{5}$	$\mathrm{a}$	$\frac{1}{3}$	$\frac{1}{5}$	$\mathrm{~b}$

$\bar{X}=2.3$

$-a+6 b=\frac{9}{10} \ldots (1)$

$\sum P_{i}=\frac{1}{5}+a+\frac{1}{3}+\frac{1}{5}+b=1$

$a+b=\frac{4}{15} \ldots (2)$

From equation $(1)$ and $(2)$

$a=\frac{1}{10}, b=\frac{1}{6}$

$\sigma^{2}=\Sigma p_{i} x_{i}^{2}-(\bar{X})^{2}$

$\frac{1}{5}(4)+a(1)+\frac{1}{3}(9)+\frac{1}{5}(16)+b(36)-(2.3)^{2}$

$=\frac{4}{5}+a+3+\frac{16}{5}+36 b-(2.3)^{2}$

$=4+a+3+36 b-(2.3)^{2}$

$=7+a+36 b-(2.3)^{2}$

$=7+\frac{1}{10}+6-(2.3)^{2}$

$=13+\frac{1}{10}-\left(\frac{23}{10}\right)^{2}$

$=\frac{131}{10}-\left(\frac{23}{10}\right)^{2}$

$=\frac{1310-(23)^{2}}{100}$

$=\frac{1310-529}{100}$

$=\frac{781}{100}$

$100 \sigma^{2}=781$

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.