Question

A university considers giving only pass/fail grades to freshmen to reduce competition and stress. The student newspaper interviews faculty members and reports their opinions of the proposed policy. Suppose that 70% of the faculty favor the pass/fail proposal. Assuming 50 faculty members are interviewed. The goal is to find the probability that a majority (26 or more) will favor the proposal.

Answer 1

0.9983 = 99.83% probability that a majority will favor the proposal.

Explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed distributions 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.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

Suppose that 70% of the faculty favor the pass/fail proposal. Assuming 50 faculty members are interviewed.

This means, respectively, that
`p = 0.7, n = 50`

Mean and standard deviation:

`\mu = E(X) = np = 50*0.7 = 35`

`\sigma = √(V(X)) = √(np(1-p)) = √(50*0.7*0.3) = 3.24`

Probability that a majority (26 or more) will favor the proposal.

Using continuity correction, this is
`P(X \geq 26 - 0.5) = P(X \geq 25.5)`, which is 1 subtracted by the pvalue of Z when X = 25.5. So

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

`Z = (25.5 - 35)/(3.24)`

`Z = -2.93`

`Z = -2.93` has a pvalue of 0.0017

1 - 0.0017 = 0.9983

0.9983 = 99.83% probability that a majority will favor the proposal.