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 σ^2, then 100σ^2 is equal to :

Correct answer: 781. Given mean is μ=∑p_ix_i=2.3 ⇒(-2)/5-a+1+4/5+6b=2.3 ⇒6b-a=0.9 ...(1) Also ∑p_i=1 ⇒1/5+a+1/3+1/5+b=1 ⇒a+b=4/15 ...(2) Solving, we get a=1/10; b=1/6 Variance is σ^2=∑p_ixi2-μ^2 ⇒σ^2=1/5(4)+1/10(1)+1/3(9)+1/5(16)+1/6(36)-5.29 ⇒σ^2=7.81 ⇒100σ^2=781

Q57 / 90 · JEE Main 2021 (01 Sep Shift 2) · Mathematics

Let $X$ be a random variable with distribution.

$x$ $- 2$ $- 1$ $3$ $4$ $6$

$P (X = x)$ $\frac{1}{5}$ $a$ $\frac{1}{3}$ $\frac{1}{5}$ $b$

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

Correct answer: 781

Solution

Given mean is $μ = \sum p_{i} x_{i} = 2.3$

$\Rightarrow \frac{- 2}{5} - a + 1 + \frac{4}{5} + 6 b = 2.3$

$\Rightarrow 6 b - a = 0.9 . . . (1)$

Also $\sum p_{i} = 1$

$\Rightarrow \frac{1}{5} + a + \frac{1}{3} + \frac{1}{5} + b = 1$

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

Solving, we get $a = \frac{1}{10}; b = \frac{1}{6}$

Variance is $σ^{2} = \sum p_{i} x_{i}^{2} - μ^{2}$

$\Rightarrow σ^{2} = \frac{1}{5} (4) + \frac{1}{10} (1) + \frac{1}{3} (9) + \frac{1}{5} (16) + \frac{1}{6} (36) - 5.29$

$\Rightarrow σ^{2} = 7.81$

$\Rightarrow 100 σ^{2} = 781$

Hence, the correct answer is 781.

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

Let X be a random variable with distribution. x -2 -1 3 4 6 P(X=x) 15 a 13 15 b If the mean of X is 2.3 and variance of X is σ2, then 100σ2 is equal to :

Solution

Relevant Questions

Let $X$ be a random variable with distribution.

$x$ $- 2$ $- 1$ $3$ $4$ $6$

$P (X = x)$ $\frac{1}{5}$ $a$ $\frac{1}{3}$ $\frac{1}{5}$ $b$

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