1. Tardigrade
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  4. If m and σ2 are the mean and variance of the random variable X, whose distribution in given by X=x 0 1 2 3 P(X=x) (1/3) (1/2) 0 (1/6) Then,

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Q. If $m$ and $\sigma^{2}$ are the mean and variance of the random variable $X$, whose distribution in given by
$X=x$ 0 1 2 3
$P(X=x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$

Then,

Probability - Part 2

Solution:

Given, distribution is
$X=x$ 0 1 2 3
$P(X=x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$

$\therefore $ Mean, $ m=\displaystyle\sum_{i=1}^{4} p_{i} x_{i} $
$=0 \times \frac{1}{3}+1 \times \frac{1}{2}+2 \times 0+3 \times \frac{1}{6}=1 $
Variance, $\sigma^{2}=\displaystyle\sum_{i=1}^{4} p_{i}\left(x_{i}-m\right)^{2}$
$=\frac{1}{3}(0-1)^{2}+\frac{1}{2}(1-1)^{2}+0(2-1)^{2}+\frac{1}{6}(3-1)^{2}=1$
$\therefore m=\sigma^{2}=1$