Question

A university administrator expects that 25% of students in a core course will receive an

Answer 1

Answer:

18.67% probability that the proportion of students that receive an a is 0.20 or less.

Step-by-step explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

$\mu = 0.25$.

The standard deviation of a proportion $\pi$ is given by the following formula.

$\sigma = \sqrt{\frac{\pi(1-\pi)}{n}}$

A university administrator expects that 25% of students in a core course will receive an a. There are 60 students. So $\pi = 0.25, n = 60$ and $\sigma = \sqrt{\frac{0.25*(0.75)}{60}} = 0.0559$

The probability that the proportion of students that receive an a is 0.20 or less is

This is the pvalue of Z when X = 0.2. So

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

$Z = \frac{0.2 - 0.25}{0.0559}$

$Z = -0.89$

$Z = -0.89$ has a pvalue of 0.1867.

So there is an 18.67% probability that the proportion of students that receive an a is 0.20 or less.

Answer 2

75% i would think if you subtract 25% from 100% of the class