Question

The Director of the Wellness Center is concerned that the number of hours of sleep that students who livc in thc dorms arc gctting is less than other university students across the nation. Bascd on rescarch from The National Council of Student Wcllncss Organizations, the Dircctor knows that the mean number of hours of sleep each night for the population of university students in the USA is 6.7 hours with a standard deviation of 1.2 hours The Director took a random sample of 50 students who live in the dorms and had them record how many hours of sleep thcy got on a randomly selected night. The results of that sample are recorded in the column titled "Hours of Sleep during Fall Semester". Use this data to answer the following questions. A) Create a histogram and a normal probability plot (QQ plot in StatCrunch) of the sample data. Does the data appear to have come from a normal distribution? Use the attributes ol the 2 graphs to support your conclusion. Include the graphs with your answer B) Using the information from The National Council of Student Wellness Organizations, calculate P75, the number of hours of sleep that 75% ofthe university students got less than in a night. You can find this value using the normal distribution C) Using the inlormation Irom The National Council of Student Wellness Organizations calculate the probability that a random sample of 50 students would have a mean number of hours of sleep less than 6.3, the mean of the sample? You can find this probability using the normal distribution and Central Limit Theorem if applicable

Hours of Sleep during Fall Semester
5.3
5.4
6.4
8.3
5.8
7.5
4.7
6.8
5.9
5.7
6.7
6.6
7.6
6.8
7.3
4.6
6.4
5.8
6.8
4.3
9.3
5.2
8.1
7.1
7.4
7.4
7.4
5.3
5.5
8.3
6.1
7.8
5.5
6.1
7.7
6.5
6.4
4.2
5.2
6.8
7.7
5.5
5.3
4.6
6.4
4.9
7
4.9
5.8
5.7

Answer 1

a)

7 15 13 11 31 0 a567891

Histogram 15 13 7 3 4-5 5-6 6-7 7-8 8-9 9-10

Z Values Plot 20.00 15.00 10.00 5.00 0.00 240560 HL 5.00 10.00 15.00

Since the shape of the above histogram is almost a bell-shaped curve, hence the data is normally distributed.

b)

Using the z-table, we get Probability ~ 0.75 at z = 0.675

Hence,

(X-Mean)/Std Error = 0.675

Mean = 6.316

Std Dev = 1.17

n = 50

Std Error = Std Dev/n^1/2 = 0.1655,

So X = 0.675 * 0.1655 + 6.316 = 6.43

75% of the university students got less than 6.43 hours of sleep.

c)

P(X<6.3) = P(z<(6.3-6.316)/0.1655) = P(z<-0.10) using the z table

= 0.4602 or 46.02%