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If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$, whose distribution is given by
Then
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Then
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Verified Answer
The correct answer is:
$m=\sigma^2=1$
Given, distribution is
$\begin{aligned}\therefore Mean, m & =\sum_{i=1}^4 p_i x_i \\ & =0 \times \frac{1}{3}+1 \times \frac{1}{2}+2 \times 0+3 \times \frac{1}{6}\end{aligned}$
$=0+\frac{1}{2}+0+\frac{1}{2}=1$
Variance, $\sigma^2=\sum_{i=1}^4 p_i\left(x_i-m\right)^2$
$\begin{aligned}
& =\frac{1}{3}(0-1)^2+\frac{1}{2}(1-1)^2 \\
& \quad+0(2-1)^2+\frac{1}{6}(3-1)^2 \\
& =\frac{1}{3}+0+0+\frac{2}{3}=1 \\
& m=\sigma^2=1
\end{aligned}$
$\therefore \quad m=\sigma^2=1$
$\begin{aligned}\therefore Mean, m & =\sum_{i=1}^4 p_i x_i \\ & =0 \times \frac{1}{3}+1 \times \frac{1}{2}+2 \times 0+3 \times \frac{1}{6}\end{aligned}$
$=0+\frac{1}{2}+0+\frac{1}{2}=1$
Variance, $\sigma^2=\sum_{i=1}^4 p_i\left(x_i-m\right)^2$
$\begin{aligned}
& =\frac{1}{3}(0-1)^2+\frac{1}{2}(1-1)^2 \\
& \quad+0(2-1)^2+\frac{1}{6}(3-1)^2 \\
& =\frac{1}{3}+0+0+\frac{2}{3}=1 \\
& m=\sigma^2=1
\end{aligned}$
$\therefore \quad m=\sigma^2=1$
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