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

Answer:

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

Step-by-step 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:

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

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.

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

In this problem, we have that:

$p = 0.7, n = 50$

So

$\mu = 0.7, \sigma = \sqrt{\frac{0.7*0.3}{50}} = 0.0648$

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

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

$Z = \frac{0.73 - 0.7}{0.0648}$

$Z = 0.46$

$Z = 0.46$ has a pvalue of 0.6772

X = 0.6

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

$Z = \frac{0.6 - 0.7}{0.0648}$

$Z = -1.54$

$Z = -1.54$ 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