Question

Suppose that 85 percent of the students at WVU live in student housing and 15 percent of students at WVU have alternative housing. If 1,776 students attending a career planning event represent a random sample from the student population, what is probability that the number of students with alternative housing will be fewer than 213?

Answer 1

P(alternate housing), p = 0.15

q = 1 - p = 0.85

Sample size, n = 1776

P(X < A) = P(Z < (A - mean)/standard deviation)

Mean = np

= 1776 x 0.15

= 266.4

Standard deviation =

=

= 15.05

P(the number of students with alternative housing will be fewer than 213) = P(X < 213)

= P(Z < (212.5 - 266.4)/15.05)                            {with continuity correction}

= P(Z < -3.58)

= 0.0002