Question

Assume the student body of a large university (with over 10,000 students) is 55% female, 45% male. Fifteen students are selected randomly. Let X represent the number of females in the sample. (4 pts) Find the mean, variance, and standard deviation of X.

Answer 1

Answer:

Mean = 8.25

Variance = 3.7125

Standard deviation = 1.927

Step-by-step explanation:

Let the probability of X be p.

Then $p=55\% = 0.55$

It follows that X is a binomial random variable.

For a binomial distribution, for n samples,

Mean, $\mu = np$

Variance, $\sigma^2 = np(1-p)$

Standard deviation, $\sigma =\sqrt{np(1-p)}$

Using values in the question, n = 15

Mean = $\mu = 15\times0.55 = 8.25$

Variance = $\sigma^2 = 15\times0.55\times(1-0.55) = 15\times0.55\times0.45 = 3.7125$

Standard deviation = $\sigma =\sqrt{3.7125} = 1.927$