After being rejected for employment, Kim Kelly learns that the Bellevue Credit Company has hired only three women among the last 19 new employees. She also learns that the pool of applicants is very large, with an approximately equal number of qualified men as qualified women.

Question

Help her address the charge of gender discrimination by finding the probability of getting three or fewer women when 19 people are hired, assuming that there is no discrimination based on gender.
(Report answer accurate to 8 decimal places).
P(at most three) =

Because this is a serious claim, we will use a stricter cutoff value for unusual events. We will use 0.5% as the cutoff value (1 in 200 chance of happening by chance). With this in mind, does the resulting probability really support such a charge?
yes, this supports a charge of gender discrimination
no, this does not support a charge of gender discrimination

Answer 1

To address the claim of gender discrimination, we can model the hiring scenario using a binomial distribution, where we have:

• \( n = 19 \) (the total number of new hires).
• \( p = 0.5 \) (the probability of hiring a woman, assuming there's no discrimination and the number of qualified men and women is approximately equal).
• \( x = \) the number of women hired, for which we will find the probability of hiring 3 or fewer women.

The binomial probability mass function is given by:

\[ P(X = k) = \binom{n}{k} p^k (1 - p)^{n-k} \]

where \( \binom{n}{k} \) is the number of combinations of \( n \) items taken \( k \) at a time.

We want to find \( P(X \leq 3) \):

\[ P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) \]

Calculating each of these probabilities:

1. \( P(X = 0) \): \[ P(X = 0) = \binom{19}{0} (0.5)^0 (0.5)^{19} = 1 \cdot 1 \cdot \left(0.5\right)^{19} = \frac{1}{524288} \approx 0.000001907 \]

2. \( P(X = 1) \): \[ P(X = 1) = \binom{19}{1} (0.5)^1 (0.5)^{18} = 19 \cdot 0.5 \cdot \left(0.5\right)^{19} = 19 \cdot \frac{1}{524288} \approx 0.000036716 \]

3. \( P(X = 2) \): \[ P(X = 2) = \binom{19}{2} (0.5)^2 (0.5)^{17} = \frac{19 \times 18}{2} \cdot (0.5)^2 \cdot \left(0.5\right)^{19} = 171 \cdot \frac{1}{524288} \approx 0.00032568 \]

4. \( P(X = 3) \): \[ P(X = 3) = \binom{19}{3} (0.5)^3 (0.5)^{16} = \frac{19 \times 18 \times 17}{6} \cdot (0.5)^3 \cdot \left(0.5\right)^{19} = 969 \cdot \frac{1}{524288} \approx 0.001847267 \]

Now, summing these probabilities:

\[ P(X \leq 3) \approx 0.000001907 + 0.000036716 + 0.00032568 + 0.001847267 \approx 0.002211570 \]

So, the probability of hiring three or fewer women is approximately:

\[ P(X \leq 3) = 0.002211570 \]

Now, let's compare this probability to the cutoff value of 0.5% (0.005):

• \( 0.002211570 < 0.005 \)

Since the calculated probability is less than the 0.5% significance level, it is indeed an unusual event.

Conclusion:

Since \( P(X \leq 3) \) is less than 0.5%, it suggests that the event of hiring three or fewer women out of 19, assuming no discrimination, is unlikely to happen by chance. Thus, the data supports the allegation of discrimination.

Therefore, the answer is:

yes, this supports a charge of gender discrimination.