SOLUTION: ch 8 12
Based on all student records at Camford University, students spend an average of 5.80 hours per week playing organized sports. The population’s standard deviation is 3.5
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Based on all student records at Camford University, students spend an average of 5.80 hours per week playing organized sports. The population’s standard deviation is 3.5
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Question 1202885: ch 8 12
Based on all student records at Camford University, students spend 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 sampling distribution of sample means. (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 your z-and standard error values to 2 decimal places. Round your intermediate values 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 your z-and standard error values to 2 decimal places. Round your intermediate values and final answer to 4 decimal places.)
You can put this solution on YOUR website! area between z-score of -1.82 and 1.82 = .9312.
that's the probabiity of getting a sample mean between 5 and 6.6 hours.
area between z-score of 0.91 and .91 = .6372.
that's the probability of getting a sample mean between 5.4 and 6.2 hours.
the following calcuilator results show these values to be true.
first you need to find the standard error.
standard error = standard devition / square root of sample size = 3.5 / sqrt(64) = .4375.
round to 2 decimal places = .44.
z-score for sample mean of 6.6 = (6.6 - 5.8) / .44 = 1.8181818.....
round to 2 decimal places = 1.82
z-score for sample mean of 5 = (5 - 5.8) / .44 = -1.8181818.....
round to 2 decimal places = -1.82
area to left of z-score of 1.82 = .9656 rounded to 4 decimal places.
area to left of z-score of -1.81 = .0344 rounded to 4 decimal places.
area in between = .9656 - .0344 = .9312
that's the probbiity of getting a sample mean between 5 and 6.6.
similar analysis was done for sample means between 5.4 and 6.2 hours.
z-scores were -.91 and .91
for example, z-score for sample mean of 5.4 was z = (5.4 - 5.8) / .44 = -.91 rounded to decimal places.
area in between was area to the left of z-score of .91 minus area to the left of z-score of -.91 which was equal to .6372.
