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 population. the scores of three students in a study group on a test are 91, 90, 92. use a sample size of 3. a. 92,90,90 x = 90.67 b. 92,90,91 x = 91 c. 91,91,90 x = 90.67 d. 91,90,91 x = 90.67 e. 90,92,91 x = 91 f. 91,92,90 x = 91 g. 90,92,92 x = 91.33 h. 90,90,91 x = 90.33 i. 91,92,91 x = 91.33 j. 92,91,91 x = 91.33 k. 92,90,92 x = 91.33 l. 92,91,90 x = 91 m. 92,91,92 x = 91.67 n. 90,91,90 x = 90.33 o. 92,92,90 x = 91.33 p. 92,92,92 x = 92 q. 91,90,90 x = 90.33 r. 90,92,90 x = 90.67 s. 90,91,91 x = 90.67 t. 91,91,91 x = 91 u. 91,92,92 x = 91.67 v. 92,92,91 x = 91.67 w. 91,90,92 x = 91 x. 90,90,92 x = 90.67 y. 90,90,90 x = 90 z. 91,91,92 x = 91.33 the mean of the sampling distribution is (round to two decimal places as needed.)

Explanation:

Step1: Calculate population mean

The population data is \(91,90,92\). The formula for the mean \(\mu\) of a population is \(\mu=\frac{\sum_{i = 1}^{N}x_{i}}{N}\), where \(N\) is the number of data - points and \(x_{i}\) are the individual data - points. Here, \(N = 3\), \(x_1=91\), \(x_2 = 90\), \(x_3=92\). So, \(\mu=\frac{91 + 90+92}{3}=\frac{273}{3}=91\).

Step2: Calculate population standard deviation

The formula for the population standard deviation \(\sigma\) is \(\sigma=\sqrt{\frac{\sum_{i = 1}^{N}(x_{i}-\mu)^{2}}{N}}\). First, \((91 - 91)^{2}=0\), \((90 - 91)^{2}=1\), \((92 - 91)^{2}=1\). Then \(\sum_{i = 1}^{3}(x_{i}-\mu)^{2}=0 + 1+1 = 2\). So, \(\sigma=\sqrt{\frac{2}{3}}\approx0.82\).

Step3: List all samples of size \(n = 3\) with replacement and their means

The population has 3 values \(\{90,91,92\}\). The total number of samples of size \(n = 3\) with replacement is \(3\times3\times3=27\).
For example, for the sample \((90,90,90)\), the sample mean \(\bar{x}=\frac{90 + 90+90}{3}=90\); for the sample \((90,90,91)\), \(\bar{x}=\frac{90 + 90+91}{3}=\frac{271}{3}\approx90.33\); for the sample \((90,91,92)\), \(\bar{x}=\frac{90 + 91+92}{3}=91\).
The mean of the sampling distribution of the sample mean \(\mu_{\bar{x}}\) is equal to the population mean.

Answer:

The mean of the sampling distribution is \(91.00\)