Suppose \(X\) has the following probability mass function \(P(X=0)=0.2, P(X=1)=0.5\), \(P(X=2)=0.3\). What…
Suppose \(X\) has the following probability mass function \(P(X=0)=0.2, P(X=1)=0.5\), \(P(X=2)=0.3\). What \(E\left[X^2\right]=\) ?
- 2.89
- 1.70
- 1.10
- 1.21
Solution
Given probability distribution is
\(\begin{array}{c|c|c|c}
\hline X & 0 & 1 & 2 \\
\hline P(X) & 0.2 & 0.5 & 0.3 \\
\hline
\end{array}\)
\(\begin{aligned}
\therefore E\left(X^2\right) & =\frac{\sum p_i x_i^2}{\sum p_i}=\frac{\left((0.2) \times 0^2\right)+\left(0.5 \times 1^2\right)+\left(0.3 \times 2^2\right)}{0.2+0.5+0.3} \\
& =\frac{0.5+1.2}{1}=1.7
\end{aligned}\)
Asked in: AP EAMCET 2020 (18 Sep Shift 1)
Practice more Probability questions on Aicharya