A television station estimates that $50 \%$ of college students watch the Super Bowl. For a sample of 200 students selected at random, what is the mean and variance of the number of students who watch this
Game?
A) Mean = 100.0, Variance = 7.07
B) Mean = 50.0, Variance = 7.07
C) Mean = 50.0, Variance = 50.00
D) Mean = 100.0, Variance = 50.00

Question

Quizplus · Accepted Answer

The mean of a binomial distribution is given by $np$ and the variance by $np(1-p)$. Here, $n = 200$ and $p = 0.5$, so the mean is $200 \times 0.5 = 100$ and the variance is $200 \times 0.5 \times 0.5 = 50$.

A Television Station Estimates That 50%50 \%50% Of College Students Watch the Super Bowl

A Television Station Estimates That $50 \%$ Of College Students Watch the Super Bowl