Question

e chairman of the statistics department in a certain college believes that 70% of the department’s graduate assistantships are given to international students. A random sample of 50 graduate assistants is taken. What is the probability that the sample proportion will NOT be between 0.60 and 0.73?

Answer 1

38.46% probability that the sample proportion will NOT be between 0.60 and 0.73

Explanation:

Problems of normally distributed samples are 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.

For a proportion p in a sample of size n, we have that
`\mu = p, \sigma = \sqrt{(p(1-p))/(n)}`

In this problem, we have that:

`p = 0.7, n = 50`

So

`\mu = 0.7, \sigma = \sqrt{(0.7*0.3)/(50)} = 0.0648`

What is the probability that the sample proportion will NOT be between 0.60 and 0.73?

This is 1 subtracted by the probability that it is between 0.6 and 0.73.

Probability it is between 0.6 and 0.73

pvalue of Z when X = 0.73 subtracted by the pvalue of Z when X = 0.6. So

X = 0.73

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

`Z = (0.73 - 0.7)/(0.0648)`

`Z = 0.46`

`Z = 0.46` has a pvalue of 0.6772

X = 0.6

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

`Z = (0.6 - 0.7)/(0.0648)`

`Z = -1.54`

`Z = -1.54` has a pvalue of 0.0618

0.6772 - 0.0618 = 0.6154

NOT be between 0.60 and 0.73?

1 - 0.6154 = 0.3846

38.46% probability that the sample proportion will NOT be between 0.60 and 0.73