Suppose you are the admissions officer at a university. Your university has room
for a first year class of 5,000. When you admit students, you know that the
probability of an admitted student accepting your offer and attending your
university is 0.75. To compensate for this, you plan to admit more than 5,000
students.
(a) If you admit 6,500 students, what is the probability that more than 5,000 will
accept your offer of admission?
(b) If you admit 6,550 students, what is the probability that more than 5,000 will
accept your offer of admission?
(c) If you want the probability that more than 5,000 will accept your offer of
admission to be 0.01, approximately how many students should you admit?

:) :) :)

For this problem, you can use the normal approximation to the binomial distribution. Then it becomes simply an exercise of calculating the z scores and using the z table to determine the areas under the normal curve.

The mean would be n*p where p=0.75, and n is the number of students.

The standard deviation would be the square root of (n*p*q) where q=1-p.

u = 0.75*5000 = 3750
s = 0.75*5000*0.25 = 30.618
x = 6500

z = (6500 – 3750)
30.618

I dont think thats right? is it?

u = n*p = 6500 * 0.75 = 4500
s = sqrt(npq) = sqrt(6500*0.75*0.25) = 34.91
z = (5000 - 4500) / 34.91

ok I did it the wrong way around... Thank you!!!:)

For part c I'm still having some trouble sorry! This is what I've done but it just doesn't look right... :(

P(>5000) = 0.0099

z = -2.33

-2.33 = (5000 – (0.75*n)
Ö(0.75*0.25*n)

0.75*0.25*n*(-2.33)2 = (5000 – 0.75*n)2

n*1.018 = 24992500 + 0.5625*n

n*1.018 – n*0.5625 = 24992500

n*0.4555 = 24992500

I think z should be +2.33 (so that the area beyond z is approx .01)

Then z = (x - u)/s

2.33 = (5000 - u)/34.91

5000 - (2.33*34.91) = u = 4918.66

then 0.75 = 4916.66/n

n = 4916.66/0.75 = 6558.2