Question

the distribution of the tuitions, fees, and room and board charges of a random sample of public 4 - year degree - granting postsecondary institutions is shown in the pie chart. make a frequency distribution for the data. use $26249.50 as the midpoint for $25,000 or more. then use the table to estimate the sample mean and the sample standard deviation of the data set. complete the frequency distribution for the data. (type integers or decimals. do not round.) class $15,000 - $17,499 $17,500 - $19,999 $20,000 - $22,499 $22,500 - $24,999 $25,000 or more x f

Explanation:

Step1: Determine mid - points for each class

The mid - point of the class $\$15,000-\$17,499$ is $\frac{15000 + 17499}{2}=\$16249.5$.
The mid - point of the class $\$17,500-\$19,999$ is $\frac{17500+19999}{2}=\$18749.5$.
The mid - point of the class $\$20,000-\$22,499$ is $\frac{20000 + 22499}{2}=\$21249.5$.
The mid - point of the class $\$22,500-\$24,999$ is $\frac{22500+24999}{2}=\$23749.5$.
The mid - point of the class $\$25,000$ or more is given as $\$26249.5$.

Step2: Fill in the frequency distribution table

From the pie - chart, the frequencies for the classes are:

Class	Mid - point ($x$)	Frequency ($f$)
$\$17,500-\$19,999$	$18749.5$	$11$
$\$20,000-\$22,499$	$21249.5$	$18$
$\$22,500-\$24,999$	$23749.5$	$10$
$\$25,000$ or more	$26249.5$	$5$

Step3: Calculate the sample mean $\bar{x}$

The formula for the sample mean of a frequency distribution is $\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}$.
$\sum_{i = 1}^{n}f_ix_i=16249.5\times9 + 18749.5\times11+21249.5\times18+23749.5\times10+26249.5\times5$
$=146245.5+206244.5 + 382491+237495+131247.5$
$=1103723.5$.
$\sum_{i=1}^{n}f_i=9 + 11+18+10+5=53$.
$\bar{x}=\frac{1103723.5}{53}\approx\$20824.97$.

Step4: Calculate the sample standard deviation $s$

First, calculate $\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2$.
For $x_1 = 16249.5$, $f_1 = 9$, $(x_1-\bar{x})^2=(16249.5 - 20824.97)^2=( - 4575.47)^2 = 20934977.92$.
$f_1(x_1-\bar{x})^2=9\times20934977.92 = 188414801.28$.
For $x_2 = 18749.5$, $f_2 = 11$, $(x_2-\bar{x})^2=(18749.5 - 20824.97)^2=( - 2075.47)^2=4307593.72$.
$f_2(x_2-\bar{x})^2=11\times4307593.72 = 47383530.92$.
For $x_3 = 21249.5$, $f_3 = 18$, $(x_3-\bar{x})^2=(21249.5 - 20824.97)^2=(424.53)^2 = 180225.72$.
$f_3(x_3-\bar{x})^2=18\times180225.72=3244062.96$.
For $x_4 = 23749.5$, $f_4 = 10$, $(x_4-\bar{x})^2=(23749.5 - 20824.97)^2=(2924.53)^2=8553877.72$.
$f_4(x_4-\bar{x})^2=10\times8553877.72 = 85538777.2$.
For $x_5 = 26249.5$, $f_5 = 5$, $(x_5-\bar{x})^2=(26249.5 - 20824.97)^2=(5424.53)^2=29425597.72$.
$f_5(x_5-\bar{x})^2=5\times29425597.72=147127988.6$.
$\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2=188414801.28+47383530.92+3244062.96+85538777.2+147127988.6$
$=471709160.96$.
The formula for the sample standard deviation is $s=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}}$, where $n=\sum_{i=1}^{n}f_i = 53$.
$s=\sqrt{\frac{471709160.96}{53 - 1}}=\sqrt{\frac{471709160.96}{52}}\approx\sqrt{9071330.02}\approx\$3011.87$.

Answer:

The sample mean is approximately $\$20824.97$ and the sample standard deviation is approximately $\$3011.87$. The completed frequency - distribution table is:

Class	Mid - point ($x$)	Frequency ($f$)
$\$17,500-\$19,999$	$18749.5$	$11$
$\$20,000-\$22,499$	$21249.5$	$18$
$\$22,500-\$24,999$	$23749.5$	$10$
$\$25,000$ or more	$26249.5$	$5$