Mathematics, 10.10.2019 22:00 scottbrandon653

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 $2.05 what 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?

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Answer from: macycj8

87.64% probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68.

91.46% probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80.

5.48% probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

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.

So, lets see each question:

What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68?

The Institute reported that entry level wages for male college graduates were $21.68 per hour, with a standard deviation of 2.30.

So $\mu = 21.68, \sigma = 2.30, n = 50, s = \frac{2.30}{\sqrt{50}} = 0.3253$

This probability is the pvalue of Z when X = 21.68 + 0.50 = 22.18 subtracted by the pvalue of Z when X = 21.68 - 0.50 = 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.

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, $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?

$\mu = 18.80, \sigma = 2.05, n = 50, s = \frac{2.05}{\sqrt{50}} = 0.29$

This probability is the pvalue of $Z = \frac{0.50}{0.29} = 1.72$ subtracted by the pvalue of $Z = -\frac{0.50}{0.29} = -1.72$ , following the same logic as the question above.

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, $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?

$\mu = 18.80, \sigma = 2.05, n = 120, s = \frac{2.05}{\sqrt{120}} = 0.1871$