Question

A group of five people are randomly selected from an office of five men and nine women. The random variable X denotes the number of women in the group selected.

What is the probability mass function and expected value of X?

Answer 1

Solution:

Population size = N = 5 + 9= 14

Number of women in the population = S = 9

Sample size = n = 5

Let X be the number of women in this sample.

So ,

X follows Hyper-geometric distribution with parameters

N = 14 , S = 9 , n = 5

The PMF of the hypergeometric distribution is

P(X = x) =     ; x = 0 , 1 2 .....,n

Using this formula ,

P(0) =   = 0.0004995

Similarly,

P(1) = 0.022477522

P(2) = 0.17982018

P(3) = 0.41958042

P(4) = 0.314685315

P(5) = 0.062937063

The PMF of X is written as

x	0	1	2	3	4	5
P(x)	0.0004995	0.022477522	0.17982018	0.41958042	0.314685315	0.062937063

Now ,

Expected value of Hyper-geometric distribution is given by

E(X) = n*[S/N] = 5 * [9/14] = 3.2143

E(X) = 3.2143