Question

Suppose that historically, 36% of applicants that are offered admittance to Georgia Southern actually enroll, while the others take offers somewhere else. If Georgia Southern will accept 9000 this coming year, what is the probability that less than 3250 will actually enroll? Use the normal approximation to the binomial.

P(X < 3250) =

Answer 1

Answer)

N = 9000

P = 0.36

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 3240

N*(1-p) = 5760

Both the conditions are met so we can use standard normal z table to estimate the probability

Z = (x - mean)/s.d

Mean = n*p = 3240

S.d = √{n*p*(1-p)} = 45.5367983064

We need to find

P(x<3250)

By continuity correction

P(x<3249.5)

Z = (3249.5 - 3240)/(45.5367983064)

Z = 0.21

From z table, P(z<0.21) = 0.5832