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 word counts of five essays are 508, 639, 552, 612, and 575. use a sample size of 2. the mean of the population is 577.20 (round to two decimal places as needed.) the standard deviation of the population is blank (round to two decimal places as needed.)

Explanation:

Step1: Recall population standard - deviation formula

The formula for the population standard deviation $\sigma=\sqrt{\frac{\sum_{i = 1}^{N}(x_{i}-\mu)^{2}}{N}}$, where $x_{i}$ are the data - points, $\mu$ is the population mean, and $N$ is the number of data - points. Here, $N = 5$, $\mu=577.20$, and the data points $x_1 = 508$, $x_2=639$, $x_3 = 552$, $x_4=612$, $x_5 = 575$.

Step2: Calculate $(x_{i}-\mu)^{2}$ for each data - point

For $x_1 = 508$: $(508 - 577.20)^{2}=(-69.20)^{2}=4788.64$
For $x_2 = 639$: $(639 - 577.20)^{2}=(61.80)^{2}=3819.24$
For $x_3 = 552$: $(552 - 577.20)^{2}=(-25.20)^{2}=635.04$
For $x_4 = 612$: $(612 - 577.20)^{2}=(34.80)^{2}=1211.04$
For $x_5 = 575$: $(575 - 577.20)^{2}=(-2.20)^{2}=4.84$

Step3: Calculate $\sum_{i = 1}^{5}(x_{i}-\mu)^{2}$

$\sum_{i = 1}^{5}(x_{i}-\mu)^{2}=4788.64 + 3819.24+635.04 + 1211.04+4.84=10458.8$

Step4: Calculate the population standard deviation

$\sigma=\sqrt{\frac{10458.8}{5}}=\sqrt{2091.76}\approx45.74$

Answer:

$45.74$