Question

The first-year retention rate is the percentage of entering freshman at a given college who return to that same college for their sophomore year. Many colleges use the first-year retention rate as one measure of their quality. Suppose that the first-year retention rate at a given college is 75% (a typical number).

a. One dormitory suite holds six first-year students. What is the probability that at least four of those students return for their sophomore year?

b. Suppose that this college admits 500 new first-year students every year, and has had a retention rate of 75% for a long time. What is the mean number of students who return for their sophomore year? What is the standard deviation for this number?

c. What is the probability that, in a given year, no more than 350 first-year students return for their sophomore year?

d. Suppose that in a given year, 410 students return for their sophomore year. Is this unusually high? Explain your answer.

Answer 1

a) One dormitory suite holds 6 first year students. what is the probability that at least four of those students return for their sophomore year?
b) suppose that this college admits 500 new first-year students every year and has had a retention rate of 75% for a long time. what is the mean number of students who return for their sophomore year? what is the standard deviation for this number?
c)what is the probability that in a given year no more than 350 first year students return for their sophomore year?
d)Suppose that in a given year, 410 students return for their sophomore year. is this unusually high? explain your answer.

Please help, I really have no idea, this stuff confuses me.

Answer 2

For a) you could calculate directly, or you could use a poisson distribution function. Let me calculate directly.

The probability that exactly n return is 6-choose-n * .75^n * .25^(6-n)

P(all) = .75^6 = .1720
P(5) = 6*(.75^5)*.25 = .3540
P(4) = ((6*5)/2)*(.75^4)*(.25^2) = .2966

So P(6,5,or4)=.1720+.3540+.2966=.8226

b) the for a binominal, the:
SD = sqrt(n*p*q) = sqrt(500*.75*.25) = 9.68

So, the expected mean is .75*500 = 375 with a SD of 9.68

c) 350 is 25 from the mean or 25/9.68 = 2.58. Looking up this value in a standard normal distribution table is .9951 Ergo, the probability of no more than 350 is 0.0049 or 0.49%

d) take in from here, follow the same logic as in c)