Question

The Office of Student Services at a large western state university maintains information on the study habits of its full-time students. Their studies indicate that the mean amount of time undergraduate students study per week is 20 hours. The hours studied follows the normal distribution with a population standard deviation of seven hours. Suppose we select a random sample of 125 current students. What is the probability that the mean of this sample is between 19.25 hours and 21.0 hours

Answer 1

There is a probability of 77% that the mean of this sample is between 19.25 hours and 21.0 hours

Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

$z=\frac{x-\mu}{\sigma/\sqrt{n} } \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation,n=sample\ size$

Given that μ = 20, σ = 7, n = 125.

For x = 19.25:

$z=\frac{19.25-20}{7/\sqrt{125} } =-1.20\\\\\\For\ x=21:\\\\z=\frac{21-20}{7/\sqrt{125} } =0.62$

From the normal distribution table, P(-1.20 < z < 0.62) = P(z < 0.62) - P(z < -1.2) = 0.8849 - 0.1151 = 0.7698 = 77%

There is a probability of 77% that the mean of this sample is between 19.25 hours and 21.0 hours

Find out more on z score at: https://brainly.com/question/25638875