Mark the correct alternative in the following question:

The probability distribution of a discrete random variable X is given below:

X: 2 3 4 5

P(X): $\frac{5}{k}$ $\frac{7}{k}$ $\frac{9}{k}$ $\frac{11}{k}$

The value of k is

(a) 8 (b) 16 (c) 32 (d) 48

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Solution

The probability distribution of a discrete random variable X is given below:

X: 2 3 4 5

P(X): $\frac{5}{k}$ $\frac{7}{k}$ $\frac{9}{k}$ $\frac{11}{k}$

$As, \sum p_{i} = 1 \Rightarrow \frac{5}{k} + \frac{7}{k} + \frac{9}{k} + \frac{11}{k} = 1 \Rightarrow \frac{32}{k} = 1 ∴ k = 32$

Disclaimer: The question is incorrect, it should be "k" instead "E(X)". The same has been corrected here.

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X:	2	3	4	5
P(X):	$\frac{5}{k}$	$\frac{7}{k}$	$\frac{9}{k}$	$\frac{11}{k}$