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. g. 508, 552, $\bar{x}=530$ h. 639, 508, $\bar{x}=573.5$ i. 552, 508, $\bar{x}=530$ j. 639, 575, $\bar{x}=607$ k. 575, 552, $\bar{x}=563.5$ l. 575, 640, $\bar{x}=607.5$ m. 575, 575, $\bar{x}=575$ n. 508, 612, $\bar{x}=560$ o. 639, 552, $\bar{x}=595.5$ p. 552, 575, $\bar{x}=563.5$ q. 575, 612, $\bar{x}=593.5$ r. 508, 575, $\bar{x}=541.5$ s. 612, 639, $\bar{x}=625.5$ t. 639, 612, $\bar{x}=625.5$ u. 639, 639, $\bar{x}=639$ v. 612, 639, $\bar{x}=559.5$ w. 552, 552, $\bar{x}=552$ x. 612, 508, $\bar{x}=560$ y. 552, 612, $\bar{x}=582$ z. 575, 639, $\bar{x}=607$ . 508, 508, $\bar{x}=508$ the mean of the sampling distribution is 577.2. (round to two decimal places as needed.) the standard deviation of the sampling distribution is. (round to two decimal places as needed)

Explanation:

Step1: Recall the formula for the standard deviation of a sampling - distribution

The formula for the standard deviation of the sampling distribution (also known as the standard error) when sampling with replacement from a population of size $N$ with population standard deviation $\sigma$ and sample size $n$ is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$. First, we need to find the population mean $\mu$ and population standard deviation $\sigma$.
The population data is $x_1 = 508,x_2=639,x_3 = 552,x_4=612,x_5 = 575$, and $N = 5$.
The population mean $\mu=\frac{\sum_{i = 1}^{N}x_i}{N}$.
$\sum_{i=1}^{5}x_i=508 + 639+552+612+575=2886$.
$\mu=\frac{2886}{5}=577.2$.

Step2: Calculate the population standard deviation $\sigma$

The formula for the population standard deviation is $\sigma=\sqrt{\frac{\sum_{i = 1}^{N}(x_i-\mu)^2}{N}}$.
$(x_1-\mu)^2=(508 - 577.2)^2=(-69.2)^2 = 4788.64$.
$(x_2-\mu)^2=(639 - 577.2)^2=(61.8)^2=3819.24$.
$(x_3-\mu)^2=(552 - 577.2)^2=(-25.2)^2 = 635.04$.
$(x_4-\mu)^2=(612 - 577.2)^2=(34.8)^2 = 1211.04$.
$(x_5-\mu)^2=(575 - 577.2)^2=(-2.2)^2 = 4.84$.
$\sum_{i = 1}^{5}(x_i - \mu)^2=4788.64+3819.24+635.04+1211.04+4.84 = 10458.8$.
$\sigma=\sqrt{\frac{10458.8}{5}}=\sqrt{2091.76}\approx45.74$.

Step3: Calculate the standard deviation of the sampling - distribution

We are given $n = 2$. Using the formula $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, substituting $\sigma\approx45.74$ and $n = 2$.
$\sigma_{\bar{x}}=\frac{45.74}{\sqrt{2}}\approx\frac{45.74}{1.414}\approx32.34$.

Answer:

$32.34$