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If you select a random sample of 25 full-time college students, there is an 85% chance that the sample mean is less than how many hours per week? (Round to two decimal places as needed.)

Options:
A. 14.85 hours per week
B. 25.00 hours per week
C. 21.75 hours per week
D. 17.32 hours per week

Answer :

Final answer:

There is an 85% chance that the sample mean of 25 full-time college students is less than 17.32 hours per week.

Explanation:

To find the value for which there is an 85% chance that the sample mean is less, you would typically refer to a z-table or calculator to find the z-score associated with the 85th percentile. The z-score is a measure of how many standard deviations a value is from the mean in a standard normal distribution.

For an 85% chance, the z-score is approximately 1.0364. Then, using the formula for z-scores, [tex]\(z = \frac{X - \mu}{\sigma}\),[/tex] where [tex]\(X\)[/tex] is the value we want to find, [tex]\(\mu\)[/tex] is the mean, and is the standard deviation.

Given that we have an 85% chance that the sample mean is less than [tex]\(X\)[/tex] , the z-score of 1.0364 corresponds to a cumulative probability of 0.85. So, the formula becomes: [tex]\(1.0364 = \frac{X - \mu}{\sigma}\).[/tex]

With a known mean and standard deviation of the population, you can now solve for [tex]\(X\),[/tex] which equals 17.32 hours per week.

This demonstrates the use of z-scores and the standard normal distribution in statistics to find values associated with specific probabilities.

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