Question

If X is a random-variable with probability distribution as given below:

X = xi :	0	1	2	3
P (X = xi) :	k	3 k	3 k	k

The value of k and its variance are

(a) 1/8, 22/27
(b) 1/8, 23/27
(c) 1/8, 24/27
(d) 1/8, 3/4

Answer 1

(d) 1/8, 3/4

We know that the sum of probabilities in a probability distribution is always 1.

∴ P (X = 0) + P (X = 1) + P (X = 2) + P (X = 3) = 1

$$\begin{gathered} \Rightarrow k+3k+3k+k=1 \\ \Rightarrow 8k=1 \\ \Rightarrow k=\frac{1}{8} \end{gathered}$$

Now,

xi	pi	pixi	pixi2
0	k = $\frac{1}{8}$	0	0
1	3k = $\frac{3}{8}$	$\frac{3}{8}$	$\frac{3}{8}$
2	3k = $\frac{3}{8}$	$\frac{6}{8}$	$\frac{12}{8}$
3	k = $\frac{1}{8}$	$\frac{3}{8}$	$\frac{9}{8}$
		${\sum }_{}^{}$pixi = $\frac{12}{8}=\frac{3}{2}$	${\sum }_{}^{}$pixi2 = $\frac{24}{8}$

$$\begin{gathered} \mathrm{Mean}=\sum p_i{x}_i=\frac{3}{2} \\ \mathrm{Variance}=\underset{}{\overset{}{\sum p_i{x_i}^2}}-{\left(\mathrm{Mean}\right)}^2=\frac{24}{8}-{\left(\frac{3}{2}\right)}^2=\frac{24}{8}-\frac{9}{4}=\frac{24-18}{8}=\frac{6}{8}=\frac{3}{4} \end{gathered}$$