22. In the Department of Education at UR University, student records suggest tha

Question

22. In the Department of Education at UR University, student records suggest that the population of students spends an average of 5.5 hours per week playing organized sports. The population's standard deviation is 2.2 hours per week. Based on a sample of 121 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. b. What is the chance HLI will find a sample mean between 5 and 6 hours? c. Calculate the probability that the sample mean will be between 5.3 and 5.7 hours. d. How strange would it be to obtain a sample mean greater than 6.5 hours?

Explanation / Answer

Average = 5.5 hours per week

stdev = 2.2 hours per week; sample size n = 121

a) std error of the sample mean = stdev/sqrt(n) = 2.2/sqrt(121) = 0.20

b) (5 < mean < 6)

= P[Z < (6 - 5.5)/0.2] - P[Z < (5 - 5.5)/0.2]

= P[Z < 2.5] - P[Z < -2.5]

= 0.9938 - 0.0062

= 0.9876

c) (5.3 < mean < 5.7)

= P[Z < (5.7 - 5.5)/0.2] - P[Z < (5.3 - 5.5)/0.2]

= P[Z < 1] - P[Z < -1]

= 0.8413 - 0.1587

= 0.6827

d) (X > 6.5)

= 1 - P[Z < (6.5 - 5.5)/0.2]

= 1 - P[Z < 5]

= 1 - 0.9999

= 0.0001

There is very little chance of getting value > 6.5 hours

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