Question

The Economic Policy Institute periodically issues reports on wages of entry level workers. The Institute reported that entry level wages for male college graduates were $21.68 per hour and for female college graduates were $18.80 per hour in 2011 (Economic Policy Institute website, March 30, 2012). Assume the standard deviation for male graduates is $2.30, and for female graduates it is $2.05What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68?What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80?What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?

Answer 1

Answer:

There is an 87.64% probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean.

There is a 91.46% probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean.

There is a 5.84% probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean.

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

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

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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

Assume the standard deviation for male graduates is $2.30, and for female graduates it is $2.05What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68?

We have that $\mu = 21.68, \sigma = 2.30$.

We have to find the standard deviation of the sample. That is:

$s = \frac{\sigma}{\sqrt{50}} = 0.3253$.

$21.68 + 0.50 = 22.18$

$21.68 - 0.50 = 21.18$

So the probability is the pvalue of $X = 22.18$ subtracted by the pvalue of $X = 21.18$

X = 22.18

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

$Z = \frac{22.18 - 21.68}{0.3253}$

$Z = 1.54$

$Z = 1.54$ has a pvalue of 0.9382.

X = 21.18

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

$Z = \frac{21.18 - 21.68}{0.3253}$

$Z = -1.54$

$Z = -1.54$ has a pvalue of 0.0618.

This means that there is a 0.9382-0.0618 = 0.8764 = 87.64% probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean.

What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80?

Assume the standard deviation for male graduates is $2.30, and for female graduates it is $2.05.

We have that $\mu = 18.80, \sigma = 2.05$.

We have to find the standard deviation of the sample. That is:

$s = \frac{\sigma}{\sqrt{50}} = 0.29$.

$18.80 + 0.50 = 19.30$

$18.80 - 0.50 = 18.30$

So the probability is the pvalue of $X = 19.30$ subtracted by the pvalue of $X = 18.30$

X = 19.30

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

$Z = \frac{19.30 - 18.80}{0.29}$

$Z = 1.72$

$Z = 1.72$ has a pvalue of 0.9573.

X = 21.18

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

$Z = \frac{18.30 - 18.80}{0.3253}$

$Z = -1.72$

$Z = -1.72$ has a pvalue of 0.0427.

This means that there is a 0.9573-0.0427 = 0.9146 = 91.46% probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean.

What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean?

This is $P(X \leq 18.50)$, with $n = 120$.

This is the pvalue of Z when $X = 18.50$

$s = \frac{\sigma}{\sqrt{120}} = 0.1871$.

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

$Z = \frac{18.50 - 18.80}{0.1871}$

$Z = -1.60$

$Z = -1.60$ has a pvalue of 0.0584.

This means that there is a 5.84% probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean.