Question

Suppose a university announced that it admitted 2,500 students for the following year's freshman class. However, the university has dorm room spots for only 1,786 freshman students. If there is a 70% chance that an admitted student will decide to accept the offer and attend this university, what is the approximate probability that the university will not have enough dormitory room spots for the freshman class?

Answer 1

There is a 5.26% probability that the university will not have enough dormitory room spots for the freshman class

Explanation:

For each admitted student, there are only two possible outcomes. Either they accept the offer, or they do not. This means that we can solve this problem using concepts the binomial probability distribution.

Binomial probability distribution:

This is the probability of x sucesses on n repeated trials, with p probability.

It has an expected value of:

`E(X) = np`

It has a standard deviation of:

`√(Var(X)) = √(np(1-p))`

We are also going to use concepts of the binomial probability distribution:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

A binomial probability distribution can be approximated to the normal using
`\mu = E(X), \sigma = √(Var(X))`

In this problem, we have that:

2500 students were admitted.

There is a 70% that they accept.

So

`E(X) = 2500*0.7 = 1750`

`√(Var(X)) = √(2500*0.7*0.3) = 22.91`

There are spots for only 1,786 students.

What is the approximate probability that the university will not have enough dormitory room spots for the freshman class?

This is the probability that there 1787 or more students admitted.

We use the binomial approximation to the normal, with
`\mu = E(X) = 1750, \sigma = √(Var(X)) = 22.91`.

So this probability is 1 subtracted by the pvalue of Z when
`X = 1787`

`Z = (X - \mu)/(\sigma)`

`Z = (1787 - 1750)/(22.91)`

`Z = 1.62`

`Z = 1.62` has a pvalue of 0.9474.

This means that there is a 1-0.9474 = 0.0526 = 5.26% probability that the university will not have enough dormitory room spots for the freshman class