A western city has 18 police officers eligible for promotion. Eleven of the 18 are Hispanic. Suppose only five of the police officers are chosen for promotion and that one is Hispanic. If the officers chosen for promotion had been selected by chance alone, what is the probability that one or fewer of the five promoted officers would have been Hispanic? What might this result indicate?
Solution
Given that $N = 18$, $A = 11$ Hispanic, $n = 5$.
The probability distribution of $X$ is Hypergeometric distribution with $N = 18$, $A = 11$, $n = 5$.
The probability that one or fewer of the five promoted officers would have been Hispanic is
$$ \begin{aligned} P(X\leq 1) &= P(X=0) + P(X=1)\\ &=\frac{\binom{11}{0}\binom{7}{5}}{\binom{18}{5}} + \frac{\binom{11}{1}\binom{7}{4}}{\binom{18}{5}} \\ &=0.0449 + 0.0025 \\ &= 0.0474 \end{aligned} $$
As the probability 0.0474 is small, it is fairly unlikely that these results occur by chance.
Further Reading
- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators
