Question

Suppose that historically, 35% of applicants that are offered admittance to Georgia Southern actually enroll, while the others take offers somewhere else. If Georgia Southern will accept 9150 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

Solution :

Given that,

p = 0.35

q = 1 - p =1--0.35=0.65

n = 9150

Using binomial distribution,

Mean = = n * p = 9150*0.35=3202.5

Standard deviation = = n * p * q =  9150*0.35*0.65=45.6248

Using continuity correction ,

P(x< 3250 ) = P((x - ) < (3249.5-3202.5) /45.6248 )

= P(z <1.03 )

Using z table

=0.8485

Probability = 0.8485