Find probability

Question

The faculty of a small eastern university is considering adding grade point average within a student's major to each transcript. You have been asked to sample students to determine student attitudes toward this proposed change. Assume that you sample, with replacement(i.e., each student is equally likely to be chosen on each draw), students at the university and that there is a ten percent chance that a student at this university is opposed to the proposed change in the transcript, independently from student to student.

a) If your sample consists of 18 individuals, find the probability that more than one individual in the sample is opposed to the proposed change.

b) If your sample consists of 64 individuals, find the expected value and the standard deviation of the number of students in the sample who favor the change in the transcript.

c) If your sample consists of 200 individuals, find the approximate probability that fewer than 10 of the students in your sample will be opposed to the proposed change in the transcript and indicate why the approximation method you use is appropriate for this problem.

Answer 1

This is an application of the bernouli distribution, also known as the binomial distribution.

Let p equal the probability that a student opposes the change. You are told that 10% of students oppose the change, so p = 0.1. That means 90% approve. Let q = 0.9, then:

P(x) = p^x q^(n-x) * n! / (x! * (n - x)!)

is the probability of x students in a sample of n opposing the change.

For the first question (more than one in a sample of 18), compute P(0) and P(1), then:

P(more than 1) = 1 - P(0) - P(1) = 0.55

For question two, use the mean of the bernouli distribution mean = np = 6.4. The variance is npq, so standard deviation is sqrt(npq) = 2.4

For question three, use the gaussian (normal) distribution approximation to the bernouli. You already have the mean (np) and variance (npq). It's justified because np is large enough.

Assuming that you're to use a table for the normal distribution, you need to standardize the variable first,

Let z = (x - np)/sqrt(npq)

This allows you to use a table of zero-mean, unit variant normal distribution to compute the cumulative density of x <= 9.

I find P(x <= 9) = 0.0048