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 population.

Population Mean = 29.33

Population Standard Deviation = 26.13

The samples of size 2 with replacement and their respective means and standard deviations

1) (66, 66)

Mean = 66.00, Standard deviation = 0.00

2) (66, 15)

Mean = 40.50. Standard deviation = 36.06

3) (66, 7)

Mean = 36.50. Standard deviation = 41.72

4) (15, 15)

Mean = 15.00. Standard deviation = 0.00

5) (15, 66)

Mean = 40.50. Standard deviation = 36.06

6) (15, 7)

Mean = 11.00. Standard deviation = 5.66

7) (7, 7)

Mean = 7.00. Standard deviation = 0.00

8) (7, 66)

Mean = 36.50. Standard deviation = 41.72

9) (7, 15)

Mean = 11, Standard deviation = 5.66

The sampling distribution of sample means is (66, 40.5, 36.5, 15, 40.5, 11, 36.5, 7, 11)

Mean of sampling distribution = 29.33

Standard deviation of sampling distribution = 18.48

The population mean and the mean of sampling distribution are both equal to 29.33

The standard deviation of the population (26.13) is higher than that of the sampling distribution (20.66).

Step-by-step explanation:

Complete 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. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. (Round to two decimal places.)

The number of dvds rented by each of the three families in the past month is 66, 15, 7. Use a sample size of 2.

Solution

For the population

Mean = (Σx)/N

Σx = Sum of all the variables = 66 + 15 + 7 = 88

N = Number of variables = 3

Mean = (88/3) = 29.33

Standard deviation =  σ = √[Σ(x - xbar)²/N]  

x = each variable

xbar = mean = 29.33

N = number of variables  = 3

Σ(x - xbar)² = (66 - 29.33)² + (15 - 29.33)² + (7 - 29.33)² = 2048.6667

σ = √(2048.6667/3) = 26.13

Noting that the for the samples,

Mean = (Σx)/N

Standard deviation =  σ = √[Σ(x - xbar)²/(N-1)]

The possible samples of size 2 possible from this distribution with replacement and their respective means and standard deviations include

(66, 66)

Mean = 66.00, Standard deviation = 0.00

(66, 15)

Mean = 40.50. Standard deviation = 36.06

(66, 7)

Mean = 36.50. Standard deviation = 41.72

(15, 15)

Mean = 15.00. Standard deviation = 0.00

(15, 66)

Mean = 40.50. Standard deviation = 36.06

(15, 7)

Mean = 11.00. Standard deviation = 5.66

(7, 7)

Mean = 7.00. Standard deviation = 0.00

(7, 66)

Mean = 36.50. Standard deviation = 41.72

(7, 15)

Mean = 11, Standard deviation = 5.66

The sampling distribution of sample means is (66, 40.5, 36.5, 15, 40.5, 11, 36.5, 7, 11)

The mean and standard deviation of sampling distribution

Mean = (66+40.5+36.5+15+40.5+11+7+36.5+11)/9 = (264/9) = 29.33

Standard Deviation = 18.48

The population mean and the mean of sampling distribution are both equal to 29.33

The standard deviation of the population (26.13) is higher than that of the sampling distribution (18.48).

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