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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions2 Marks
MCQ
The mean and standard deviation of random variable $X$ are 10 and 5 respectively, then $\mathrm{E}\left(\frac{\mathrm{X}-15}{5}\right)^2=$ ________________ .
  • A
    4
  • B
    3
  • ✓
    2
  • D
    5
✓

Answer

Correct option: C.
2
(C)
$ \operatorname{Var}(X)=\sigma^2=5^2=25$
$\begin{aligned} & \operatorname{Var}(X)=E\left(X^2\right)-[E(X)]^2 \\ & \Rightarrow 25=E\left(X^2\right)-10^2 \\ & \Rightarrow E\left(X^2\right)=125 \\ & E\left(\frac{X-15}{5}\right)^2=E\left(\frac{X^2-30 X+225}{25}\right)\end{aligned}$
$\begin{aligned} & =\frac{1}{25}\left[\mathrm{E}\left(\mathrm{X}^2\right)-30 \mathrm{E}(\mathrm{X})+225\right] \\ & =\frac{1}{25}(125-300+225)=2\end{aligned}$

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