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 5 hours with a standard deviation of 3 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. thr
(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.) 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 thours (etable or technology. Round your five decimal places.)

Answer 1

Answer:

5 ; 0.728 ; 0.00002

Step-by-step explanation:

According to the Central limit theorem, for a normal distribution, the population mean equals the sample mean., Hence the population mean, μ = sample mean = 5

The standard deviation of sampling distribution :

Standard deviation / sqrt(n)

σ / sqrt(n)

σ = 3 ; n = 17

3 / sqrt(17)

3 / 4.1231056

= 0.7276

= 0.728

3.)

P(x < 2) :

Using :

Z = (x - mean) ÷ s/sqrt(n)

P(x < 2) = (2 - 5) ÷ 3/sqrt(17)

P(x < 2) = -3 / 0.7276068

P(x < 2) = - 4.1231060

Z < - 4.123 = 0.00002 (Z probability calculator).