Question

After being rejected for employment, Kim Kelly learns that the Bellevue Credit Company has hired only five women among the last 18 new employees. She also learns that the pool of applicants is very large, with an approximately equal number of qualified men as qualified women. Help her address the charge of gender discrimination by finding the probability of getting five or fewer women when 18 people are hired, assuming that there is no discrimination based on gender.
Does the resulting probability really support such a charge?

Answer 1

Answer:

0.0481 = 4.81% probability of getting five or fewer women when 18 people are hired. Since this probability is lower than 5%, it means that five or fewer woman being hired is an unlikely event, and supports the charge.

Step-by-step explanation:

For each new employee hired, there are only two possible outcomes. Either it is a woman, or it is not. The probability of an employee being a woman is independent of any other employee. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

She also learns that the pool of applicants is very large, with an approximately equal number of qualified men as qualified women.

This means that $p = 0.5$

Help her address the charge of gender discrimination by finding the probability of getting five or fewer women when 18 people are hired, assuming that there is no discrimination based on gender.

This is $P(X \leq 5)$ when $n = 18$

So

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)$

In which

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{18,0}.(0.5)^{0}.(0.5)^{18} \approx 0$

$P(X = 1) = C_{18,1}.(0.5)^{1}.(0.5)^{17} = 0.0001$

$P(X = 2) = C_{18,2}.(0.5)^{2}.(0.5)^{16} = 0.005$

$P(X = 3) = C_{18,3}.(0.5)^{3}.(0.5)^{15} = 0.0031$

$P(X = 4) = C_{18,4}.(0.5)^{4}.(0.5)^{14} = 0.0117$

$P(X = 5) = C_{18,5}.(0.5)^{5}.(0.5)^{13} = 0.0327$

$P(X \leq 5) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 0 + 0.0001 + 0.0005 + 0.0031 + 0.0117 + 0.0327 = 0.0481$

0.0481 = 4.81% probability of getting five or fewer women when 18 people are hired. Since this probability is lower than 5%, it means that five or fewer woman being hired is an unlikely event, and supports the charge.