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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions2 Marks
MCQ
The probability distribution of a discrete random variable X is given by
X-1012
P(X)$\frac{1}{3}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{3}$
Then, the value of $6 \mathrm{E}\left(\mathrm{X}^2\right)-\operatorname{Var}(\mathrm{X})$ is
  • A
    $\frac{12}{113}$
  • ✓
    $\frac{113}{12}$
  • C
    $\frac{19}{12}$
  • D
    $\frac{1}{2}$
✓

Answer

Correct option: B.
$\frac{113}{12}$
(B)
$ \mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)=-\frac{1}{3}+0+\frac{1}{6}+\frac{2}{3}=\frac{1}{2}$
$\begin{aligned} & \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ & =\frac{(-1)^2}{3}+0+\frac{1^2}{6}+\frac{2^2}{3}-\left(\frac{1}{2}\right)^2 \\ & =\frac{1}{3}+\frac{1}{6}+\frac{4}{3}-\frac{1}{4} \\ & =\frac{11}{6}-\frac{1}{4}=\frac{19}{12} \\  \therefore & 6 \mathrm{E}\left(\mathrm{X}^2\right)-\operatorname{Var}(\mathrm{X}) \\ & =6\left(\frac{11}{6}\right)-\frac{19}{12} \\ & =11-\frac{19}{12} \\ & =\frac{113}{12}\end{aligned}$

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