Answer to Question #317640 in Statistics and Probability for Kim

Question

Dan works in an insurance company. Last January, he was able to insure 2 persons. Last February, he was able to insure 3 persons. Last March, he was able to insure 5 persons. Assume that the samples of size 2 are randomly selected with replacement. Find the following:

a. The population mean

b. The population variance

c. The population standard deviation

d. The mean of the sampling distribution of means

e. The standard deviation of the sampling distribution of the sample means

Answer 1

a. population mean

$\frac{2+3+5}{3}=\frac{10}{3}=3.33$

b. population variance

$=\frac{(2-3.33)^2+(3-3.33)^2+(5-3.33)^2}{3}=\frac{4.6667}{3}=1.556$

c. population standard deviation

$=\sqrt{1.556}=1.247$

d. mean of sampling distribution

sampling mean = 3.33

e. standard deviation of sample distribution

$\sigma=\sqrt{11.8888-3.333^2}=\sqrt{0.780}=0.883$