MCQ
The probability distribution of a random variable X is given below.
X = k01234
P(X = k)0.10.40.30.20
The variance of X is
  • A
    1.6
  • B
    0.24
  • 0.84
  • D
    0.75

Answer

Correct option: C.
0.84
(C)
$\mathrm{E}(\mathrm{X})=\sum x_{\mathrm{i}} \cdot \mathrm{P}\left(x_{\mathrm{i}}\right)$
$\begin{aligned}&=0(0.1)+1(0.4)+2(0.3)+3(0.2)+4(0) \\ &=0+0.4+0.6+0.6+0 \\ &=1.6 \\ & \text { Variance }=\mathrm{E}\left(\mathrm{X}^2\right)-[\mathrm{E}(\mathrm{X})]^2 \\ &= 0^2(0.1)+1^2(0.4)+2^2(0.3)+3^2(0.2)+4^2(0)-1.6^2 \\ &= 0+0.4+1.2+1.8-2.56 \\ &= 0.84\end{aligned}$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free