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. then use the table to estimate the sample mean and the sample standard deviation of the data set. use $26249.50 as the midpoint for $25,000 or more. the sample mean is $\bar{x} = $20898.62$ (round to the nearest cent.) the sample standard deviation is $s = $ (round to the nearest cent.)

Explanation:

Step1: Organize data into a table

Class Interval	Frequency ($f$)	Mid - point ($x$)	$f\times x$	$f\times x^{2}$
$17500 - 19999$	$11$	$18749.5$	$11\times18749.5=206244.5$	$11\times(18749.5)^{2}=11\times351543250.25 = 3866975752.75$
$20000 - 22499$	$18$	$21249.5$	$18\times21249.5 = 382491$	$18\times(21249.5)^{2}=18\times451549750.25 = 8127895504.5$
$22500 - 24999$	$13$	$23749.5$	$13\times23749.5=308743.5$	$13\times(23749.5)^{2}=13\times564035250.25 = 7332458253.25$
$25000$ or more	$5$	$26249.5$	$5\times26249.5 = 131247.5$	$5\times(26249.5)^{2}=5\times688034750.25 = 3440173751.25$

The sum of frequencies $\sum f=10 + 11+18+13+5=57$.
The sum of $f\times x$ is $\sum(f\times x)=162495+206244.5+382491+308743.5+131247.5 = 1191221.5$.
The sum of $f\times x^{2}$ is $\sum(f\times x^{2})=2640477502.5+3866975752.75+8127895504.5+7332458253.25+3440173751.25=25407980764.25$.

Step2: Calculate the sample standard - deviation

The formula for the sample standard deviation $s=\sqrt{\frac{\sum(f\times x^{2})}{n - 1}-\frac{(\sum(f\times x))^{2}}{n(n - 1)}}$, where $n=\sum f = 57$.

First term: $\frac{\sum(f\times x^{2})}{n - 1}=\frac{25407980764.25}{56}\approx453713942.22$.

Second term: $\frac{(\sum(f\times x))^{2}}{n(n - 1)}=\frac{(1191221.5)^{2}}{57\times56}=\frac{141903674862.25}{3192}\approx44456038.5$.

Then $s=\sqrt{453713942.22-44456038.5}=\sqrt{409257903.72}\approx20230.12$.

Answer:

$20230.12$