Q79 / 160 · TS EAMCET 2024 (11 May Shift 1) · Mathematics
If a random variable $X$ has the following probability distribution, then its variance is
$\begin{array}{|l|c|c|c|c|}
\hline X=x & 1 & 3 & 5 & 2 \\
\hline P(X=x) & 3 K^2 & K & K^2 & 2 K \\
\hline
\end{array}$
- A$\frac{9}{4}$
- B$\frac{25}{8}$
- C$\frac{27}{16}$
- Correct answerD$\frac{15}{16}$
Solution
$\begin{aligned} & 3 k^2+k+k^2+2 k=1 \\ \Rightarrow & (4 k-1)(k+1)=0 \Rightarrow k=\frac{1}{4}(k \neq-1)\end{aligned}$
$\begin{array}{|l|c|c|c|c|}
\hline X=x & 1 & 3 & 5 & 2 \\
\hline \boldsymbol{P}(X=x) & \frac{3}{16} & \frac{1}{4} & \frac{1}{16} & \frac{1}{2} \\
\hline
\end{array}$
$\begin{aligned} & \mu=\Sigma P_i x_i=\frac{3}{16}+\frac{3}{4}+\frac{5}{16}+1=\frac{9}{4} \\ & \Sigma P_i x_i^2=\frac{3}{16}+\frac{9}{4}+\frac{25}{16}+\frac{4}{2}=\frac{3+36+25+32}{16}=6 \\ & \sigma^2=\Sigma P_i x_i^2-\mu^2=6-\frac{81}{16}=\frac{15}{16} . \end{aligned}$
$\begin{aligned} & \mu=\Sigma P_i x_i=\frac{3}{16}+\frac{3}{4}+\frac{5}{16}+1=\frac{9}{4} \\ & \Sigma P_i x_i^2=\frac{3}{16}+\frac{9}{4}+\frac{25}{16}+\frac{4}{2}=\frac{3+36+25+32}{16}=6 \\ & \sigma^2=\Sigma P_i x_i^2-\mu^2=6-\frac{81}{16}=\frac{15}{16} . \end{aligned}$
Hence, the correct answer is option: 'D'.