# Probability Question - First year class problem

So here is the problem.

The admission's office of a small university is asked to accept deposits from a number of qualified prospective first-year students so that, with probability 0.95, the size of the first year class will be less than or equal to 120. Suppose the applicants constitute a random sample from a population of applicants, 80% of whom would actually enter the first year class if accepted.

A) How many deposits should the admissions counselor accept?

B) if applicants in the number determined in part a are accepted, what is the probability that the class size will be less than 105.

I am really lost and have no idea how to solve this question. However, I do not want the answer but I do want to know how to go about solving a question like this. Does anyone have any advice for me? Thanks

• Do you know the expression for the probability that n students accept if m are offered (based on independent decisions with an 80% acceptance rate)? Oct 12, 2017 at 19:52
• Well I know that p(x <= 120) = 0.95 for the classroom size and that if 80% of students actually enter the class after they are accepted that could probably be modeled with a binomial distribution. So then is the problem simply finding the value of n representing how many people the admissions office needs to accept so that p(x <= 120) and the probability of each one person going into the class is p = 0.8? Oct 12, 2017 at 19:58
• @Phin46 please see tag your post self-study and read the site rules around posting for homework help. Oct 12, 2017 at 20:04
• This is for my stats homework, I am not looking for a free answer I just want some insight. Oct 12, 2017 at 20:04
• yes a binomial - exactly. so p(n) = ${{m}\choose{n}} \cdot 0.8^n 0.2^{m-n}$ correct? Oct 12, 2017 at 20:05

If you have never seen the probability function for a binomial, and this is a first year stats course (where sums/integrals of random variables are uncommon), I am going to assume they want you to use the normal approximation to the binomial, which in that case, let $W = \sum_{i=1}^N X_i$

where each $X_i$ are independent and identically distributed (i.i.d/random sample) Bernoulli (p=0.80) random variables. That is, $X_i = 0$ with probability 0.20, and $X_i = 1$ with probability 0.80. Also, $N$ represents the amount of people the school accepts (but not necessarily the class size, of course)

Note that W is therefore a random variable that represents the number of amount of people who will actually enter the first year class if their deposit is taken/they are accepted.

So, because the $X_i's$ are i.i.d then we know these facts:

$$E[W] = N *E[X] = N *p = 0.80N = u_W$$ $$Var[W] = N * p * (1-p) =0.16N = \sigma^2_W$$

(I changed the notation as I know in most first year stats courses they do not really talk about expectation and variance directly.)

which is exactly what you have stated in the comments.

So, we want to find N that satisfies $P(W \le 120)=0.95.$

Since $\frac{W-u_W}{\sigma_W} = Z$ where $Z \sim^{\text{approximately}}N(0,1)$,

we can manipulate the above equation into this:

$$W = Z\sigma_w + u_w$$

Then

$$P(W \le 120) = P(Z \le \frac{120-u_W}{\sigma_W})$$

I'm assuming you know how to find N from here (hint: normal table, and substitute...you probably have preset formulas for the so called normal approximation to the binomial distribution). You may also have to use the continuity correction, which would mean

$P( W \le 120) \approx P(Z \le \frac{120.5 - u_W}{\sigma_W})$

$$P(W \le 105)$$
where n is found from above. Note that once we know N, we have specific values for $U_W$ and $\sigma_W$ which means we can simply find the approximate probability by repeating calculations similar to a) but instead finding a probability rather than N.