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.

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

Step-by-step 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:

The standard deviation of the binomial distribution is:

Normal probability distribution

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

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

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 , .

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

This means, respectively, that

Mean and standard deviation:

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

Using continuity correction, this is , which is 1 subtracted by the pvalue of Z when X = 25.5. So

has a pvalue of 0.0017

1 - 0.0017 = 0.9983

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

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