Question

In the Department of Education at UR University, student records suggest that the population of students spends an average of 5.80 hours per week playing organized sports. The population's standard deviation is 3.50 hours per week. Based on a sample of 64 students, Healthy Lifestyles Incorporated (HLI) would like to apply the central limit theorem to make various estimates.

a. Compute the standard error of the sample mean. (Round your answer to 2 decimal places.)

b. What is the chance HLI will find a sample mean between 5 and 6.6 hours? (Round z and standard error values to 2 decimal places and final answer to 4 decimal places.)

c. Calculate the probability that the sample mean will be between 5.4 and 6.2 hours. (Round z and standard error values to 2 decimal places and final answer to 4 decimal places.)

d. How likely would it be to obtain a sample mean greater than 7.80 hours?

Answer #1

Solution:

Mean = 5.8, standard devaition = 3.5

No. Of sample = 64

Solution(a)

Standard error = Standard deviation/sqrt(n)

Standard error = 3.50/sqrt(64) = 0.4375 or 0.44

Solution(b)

P(5<Xbar<6.6) =p(Xbar<6.6)--(Xbar<5)

Z = (6.6-5.8)/0.44

Z = 1.82

Z =(5-5.8)/0.44 = -1.82

From z table we found

p(5<Xbar<6.6) =0.9656-0.0344 = 0.9312

Solution(c)

P(5.4<Xbar<6.2) =p(Xbar<6.2)-p(Xbar<5.4)

Z = (6.2-5.8)/0.44 = 0.91

Z =5.4-5.8/0.44 =-0.91

P(5.4<Xvar<6.2) = 0.8186-0.1814 = 0.6372

Solution(d)

P(Xbar>7.8) =1-P(Xbar<7.8)

Z =(7.8-5.8)/0.44 = 4.55

From z table we found p-value

P(Xbar>7.8) =1- 0.999996602 = 0.000003398

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