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Probability Distributions — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsProbability Distributions2 Marks
MCQ
If X is a random variable with probability distribution as given below:
X0123
P(X)k3k3kk
Then, the variance of X is
  • A
    $\frac{1}{8}$
  • B
    $\frac{23}{7}$
  • C
    $\frac{24}{27}$
  • ✓
    $\frac{3}{4}$
✓

Answer

Correct option: D.
$\frac{3}{4}$
(D)
The sum of all the probabilities in a probability distribution is always unity.
$ \therefore \quad \mathrm{k}+3 \mathrm{k}+3 \mathrm{k}+\mathrm{k}=1$
$\begin{aligned} & \Rightarrow 8 \mathrm{k}=1 \\& \Rightarrow \mathrm{k}=\frac{1}{8} \\& \mathrm{E}(\mathrm{X})-\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right) \\& =0\left(\frac{1}{8}\right)+1\left(\frac{3}{8}\right)+2\left(\frac{3}{8}\right)+3\left(\frac{1}{8}\right)=\frac{3}{2} \\& \operatorname{Var}(\mathrm{X})=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\& =0^2\left(\frac{1}{8}\right)+1^2\left(\frac{3}{8}\right)+2^2\left(\frac{3}{8}\right)+3^2\left(\frac{1}{8}\right)-\left(\frac{3}{2}\right)^2 \\& =\frac{3}{4}\end{aligned}$

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