For the following situation, find the mean and standard deviation of the populat

Question

For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the mean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the populations. The scores of three students in a study group on a test are 93, 92, 94. Use a sample size of 3. The mean of the population is (Round to two decimal places as needed.) The standard deviation of the population is (Round to two decimal places as needed.) A. 94, 94, 93 x = 93.67 B. 92, 92, 94 x = 92.67 C. 94, 92, 92 x = 92.67 D. 94, 94, 94 x = 94 E. 94, 93, 94 x = 93.67 F. 94, 92, 94 x = 93.33 G. 92, 94, 92 x = 92.67 H. 92, 92, 93 x = 92.33 I. 93, 92, 92 x = 92.33 J. 92, 93, 92 x = 92.33 K. 93, 94, 94 x = 93.67 L. 93, 92, 93 x = 92.67 M. 93, 93, 93 x = 93 N. 92, 94, 93 x = 93 O. 93, 94, 92 x = 93 P. 92, 94, 94 x = 93.33 Q. 92, 93, 94 x = 93 R. 93, 94, 93 x = 93.33 S. 94, 93, 93 x = 93.33 T. 94, 92, 93 x = 93 U. 93, 92, 94 x = 93 V. 93, 93, 94 x = 93.33 W. 93, 93, 92 x = 92.67 X. 94, 94, 92 x = 93.33 Y. 92, 93, 93 x = 92.67 Z. 92, 92, 92 x = 92 94, 93, 92 x = 93 The mean of the sampling distribution is

Explanation / Answer

Population mean = (93+92+94) / 3 = 93

Standard deviation of population =sqrt[ ( 93 - 93)^2 + ( 92 - 93)^2 + ( 94 -93)^2 / 3] = 0.8165

mean of sampling distribution = mean of population = 93

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