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

पर्याय

• 1/8, 22/27

• 1/8, 23/27

•  1/8, 24/27

• 1/8, 3/4

   

Answer 1

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

\[\Rightarrow k + 3k + 3k + k = 1\]
\[ \Rightarrow 8k = 1\]
\[ \Rightarrow k = \frac{1}{8}\]

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 = \[\frac{1}{8}\]	\[\frac{3}{8}\]	\[\frac{9}{8}\]
		\[\sum\nolimits_{}^{}\]pixi = \[\frac{12}{8} =\frac{3}{2}\]	\[\sum\nolimits_{}^{}\] pixi2 =  \[\frac{24}{8}\]

\[\text{ Mean }  = \sum p_i x_i = \frac{3}{2}\]
\[\text{ Variance } = \sum p_i {x_i}2^{}_{} - \left( \text{ Mean }  \right)^2 = \frac{24}{8} - \left( \frac{3}{2} \right)^2 = \frac{24}{8} - \frac{9}{4} = \frac{24 - 18}{8} = \frac{6}{8} = \frac{3}{4}\]