If the probability distribution of a random variable $X$ is given by
If the probability distribution of a random variable $X$ is given by
- 3
- $\frac{9}{4}$
- $\frac{3}{2}$
- $\frac{3}{4}$
Solution
Variance $=E\left(X^2\right)-(E(X))^2$
$
\begin{aligned}
& \because \quad \text { Sum of } P\left(x=x_i\right)=1 \\
& \Rightarrow \quad \frac{1}{8}+\frac{3}{8}+3 k+k=1 \Rightarrow k=\frac{1}{8}
\end{aligned}
$
and the variance
$
\begin{aligned}
=[( & \left.\left.0^2 \times \frac{1}{8}\right)+\left(1^2 \times \frac{3}{8}\right)+\left(2^2 \times \frac{3}{8}\right)+\left(3^2 \times \frac{1}{8}\right)\right] \\
& -\left[\left(0 \times \frac{1}{8}\right)+\left(1 \times \frac{3}{8}\right)+\left(2 \times \frac{3}{8}\right)+\left(3 \times \frac{1}{8}\right)\right]^2 \\
= & {\left[\frac{3}{8}+\frac{12}{8}+\frac{9}{8}\right]-\left[\frac{3}{8}+\frac{6}{8}+\frac{3}{8}\right]^2 } \\
= & \frac{24}{8}-\left(\frac{12}{8}\right)^2=3-\left(\frac{3}{2}\right)^2=3-\frac{9}{4}=\frac{12-9}{4}=\frac{3}{4}
\end{aligned}
$
Asked in: AP EAMCET 2018 (22 Apr Shift 2)
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