The Economic Policy Institute periodically issues reports on workers' wages. Suppose the institute reported that mean wages for male college

Question

graduates were $37.39 per hour and for female college graduates were $27.83 per hour in 2017. Assume the standard deviation for male
graduates is $4.60 and for female graduates it is $4.10. (Round your answers to four decimal places.)
(a) What is the probability that a sample of 60 male graduates will provide a sample mean within $1.00 of the population mean, $37.39?

Answer 1

To solve this problem, we can use the concept of sampling distribution and the Central Limit Theorem.

According to the Central Limit Theorem, as long as the sample size is sufficiently large (typically n > 30), the distribution of sample means will be approximately normal, regardless of the shape of the population distribution.

The mean of the sampling distribution of sample means is equal to the population mean, which is $37.39. The standard deviation of the sampling distribution, also known as the standard error, can be calculated using the formula:

Standard Error = σ / sqrt(n)

where σ is the population standard deviation and n is the sample size.

In this case, the population standard deviation for male graduates is $4.60, and the sample size is 60. Therefore, the standard error is:

Standard Error = 4.60 / sqrt(60) = 0.5937 (approx.)

To find the probability that a sample mean is within $1.00 of the population mean, we need to calculate the z-scores corresponding to the lower and upper limits of the range.

Lower limit: $37.39 - $1.00 = $36.39
Upper limit: $37.39 + $1.00 = $38.39

To calculate the z-scores, we use the formula:

z = (x - μ) / σ

where x is the given value, μ is the population mean, and σ is the standard deviation.

For the lower limit, the z-score is:

z1 = ($36.39 - $37.39) / 0.5937 = -1.6848 (approx.)

For the upper limit, the z-score is:

z2 = ($38.39 - $37.39) / 0.5937 = 1.6848 (approx.)

Next, we need to find the probabilities associated with these z-scores using a standard normal distribution table or a calculator. Since we are interested in the probability that the sample mean is within the specified range, we need to calculate the area under the curve between these two z-scores.

Using a standard normal distribution table or a calculator, the probability corresponding to a z-score of -1.6848 is approximately 0.0447. Similarly, the probability corresponding to a z-score of 1.6848 is also approximately 0.0447.

Finally, to find the probability that a sample of 60 male graduates will provide a sample mean within $1.00 of the population mean, we subtract the probability of being outside this range from 1:

Probability = 1 - (0.0447 + 0.0447) = 0.9106

Therefore, the probability is approximately 0.9106.