Question

Consider a small population consisting of the 100 students enrolled in an introductory statistics course. Students in the class completed a survey on academic procrastination. The average number of hours spent procrastinating when they should be studying, per exam, by all students in this course is 4 hours with a standard deviation of 2 hours. The distribution of amount of time students spend procrastinating is known to be normal. (a) Identify the value (in hours) of the population mean. 4 Correct: Your answer is correct. hr (b) Calculate the standard deviation (in hours) of the sampling distribution of the sample mean for a sample of size 17 drawn from this population. (Round your answer to three decimal places.) .485 Correct: Your answer is correct. hr (c) What is the probability of obtaining a sample mean of less than 2 hours based on a sample of size 17 if the population mean truly is 4 hours? (Use a table or technology. Round your answer to five decimal places.)

Answer 1

Solution:

Given: The average number of hours spent procrastinating when they should be studying, per exam, by all students in this course is 4 hours with a standard deviation of 2 hours.

That is: and

Part a)

the value (in hours) of the population mean= = 4

Part b)

the standard deviation (in hours) of the sampling distribution of the sample mean for a sample of size 17 drawn from this population.

Part c) What is the probability of obtaining a sample mean of less than 2 hours based on a sample of size 17 if the population mean truly is 4 hours?

Find z score:

Thus we get:

Use Excel command:

=NORM.S.DIST(z , cumulative)

=NORM.S.DIST(-4.12,TRUE)

=0.00002

Thus