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

Answer:

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:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

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:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

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:

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

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{50*0.7*0.3} = 3.24$

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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{25.5 - 35}{3.24}$

$Z = -2.93$

$Z = -2.93$ has a pvalue of 0.0017

1 - 0.0017 = 0.9983

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