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.
a. 5/5, 508, x = 541.5
b. 612, 612, x = 612
c. 612, 5/5, x = 593.5
d. 612, 552, x = 582
e. 552, 639, x = 595.5
f. 508, 639, x = 573.5
g. 508, 552, x = 530
h. 639, 508, x = 573.5
i. 552, 508, x = 530
j. 639, 575, x = 607
k. 575, 552, x = 563.5
l. 575, 640, x = 607.5
m. 575, 575, x = 575
n. 508, 612, x = 560
o. 639, 552, x = 595.5
p. 552, 575, x = 563.5
q. 575, 612, x = 593.5
r. 508, 575, x = 541.5
s. 612, 639, x = 625.5
t. 639, 612, x = 625.5
u. 639, 639, x = 639
v. 612, 639, x = 559.5
w. 552, 552, x = 552
x. 612, 508, x = 560
y. 552, 612, x = 582
z. 575, 639, x = 607
. 508, 508, x = 508
the mean of the sampling distribution is (round to two decimal places as needed.)

Explanation:

Step1: Calculate population mean

The population data is \(x_1 = 508,x_2=639,x_3 = 552,x_4=612,x_5 = 575\). The population mean \(\mu\) is given by \(\mu=\frac{\sum_{i = 1}^{n}x_i}{n}\), where \(n = 5\).
\(\mu=\frac{508 + 639+552+612+575}{5}=\frac{2886}{5}=577.2\)

Step2: Calculate population standard - deviation

First, find the squared - differences \((x_i-\mu)^2\):
\((508 - 577.2)^2=( - 69.2)^2 = 4788.64\)
\((639 - 577.2)^2=(61.8)^2 = 3819.24\)
\((552 - 577.2)^2=( - 25.2)^2 = 635.04\)
\((612 - 577.2)^2=(34.8)^2 = 1211.04\)
\((575 - 577.2)^2=( - 2.2)^2 = 4.84\)
The population variance \(\sigma^{2}=\frac{\sum_{i = 1}^{n}(x_i-\mu)^2}{n}=\frac{4788.64 + 3819.24+635.04+1211.04 + 4.84}{5}=\frac{10458.8}{5}=2091.76\)
The population standard - deviation \(\sigma=\sqrt{2091.76}\approx45.74\)

Step3: List all samples of size \(n = 2\) with replacement and their means

The number of samples of size \(n = 2\) with replacement from a population of size \(N = 5\) is \(N^n=5^2 = 25\).
The samples and their means are:
\((508,508)\), \(\bar{x}=\frac{508 + 508}{2}=508\)
\((508,552)\), \(\bar{x}=\frac{508+552}{2}=530\)
\((508,612)\), \(\bar{x}=\frac{508 + 612}{2}=560\)
\((508,575)\), \(\bar{x}=\frac{508+575}{2}=541.5\)
\((508,639)\), \(\bar{x}=\frac{508 + 639}{2}=573.5\)
\((552,508)\), \(\bar{x}=\frac{552 + 508}{2}=530\)
\((552,552)\), \(\bar{x}=\frac{552+552}{2}=552\)
\((552,612)\), \(\bar{x}=\frac{552 + 612}{2}=582\)
\((552,575)\), \(\bar{x}=\frac{552+575}{2}=563.5\)
\((552,639)\), \(\bar{x}=\frac{552 + 639}{2}=595.5\)
\((612,508)\), \(\bar{x}=\frac{612 + 508}{2}=560\)
\((612,552)\), \(\bar{x}=\frac{612 + 552}{2}=582\)
\((612,612)\), \(\bar{x}=\frac{612+612}{2}=612\)
\((612,575)\), \(\bar{x}=\frac{612 + 575}{2}=593.5\)
\((612,639)\), \(\bar{x}=\frac{612 + 639}{2}=625.5\)
\((575,508)\), \(\bar{x}=\frac{575 + 508}{2}=541.5\)
\((575,552)\), \(\bar{x}=\frac{575 + 552}{2}=563.5\)
\((575,612)\), \(\bar{x}=\frac{575 + 612}{2}=593.5\)
\((575,575)\), \(\bar{x}=\frac{575+575}{2}=575\)
\((575,639)\), \(\bar{x}=\frac{575 + 639}{2}=607\)
\((639,508)\), \(\bar{x}=\frac{639 + 508}{2}=573.5\)
\((639,552)\), \(\bar{x}=\frac{639 + 552}{2}=595.5\)
\((639,612)\), \(\bar{x}=\frac{639 + 612}{2}=625.5\)
\((639,575)\), \(\bar{x}=\frac{639 + 575}{2}=607\)
\((639,639)\), \(\bar{x}=\frac{639+639}{2}=639\)

Step4: Calculate the mean of the sampling distribution

The mean of the sampling distribution \(\mu_{\bar{x}}\) is given by \(\mu_{\bar{x}}=\frac{\sum_{i = 1}^{N^n}\bar{x}_i}{N^n}\)
\(\sum_{i = 1}^{25}\bar{x}_i=508+530+560+541.5+573.5+530+552+582+563.5+595.5+560+582+612+593.5+625.5+541.5+563.5+593.5+575+607+573.5+595.5+625.5+607+639 = 14430\)
\(\mu_{\bar{x}}=\frac{14430}{25}=577.2\)

Answer:

The mean of the population is \(577.2\), the standard - deviation of the population is approximately \(45.74\), and the mean of the sampling distribution is \(577.2\)