Question

A study reports that 36% of companies in a Country A have three or more female board directors. Suppose you select a random sample of 100 respondents.

What is the probability that the sample will have between 29% and 38% of companies in Country A that have three or more female board directors?

Answer 1

Solution:

Given:

p = proportion of companies in a Country A have three or more female board directors = 36% = 0.36

n = sample size = 100

We have to find:

Find z score:

and

Thus we get:

Look in z table for z = 0.4 and 0.02 as well as for z = -1.4 and 0.06 and find area.

P( Z< 0.42 ) =0.6628

and

P( Z < -1.46 ) =0.0721

Thus

Thus  the probability that the sample will have between 29% and 38% of companies in Country A that have three or more female board directors is 0.5907.