the distribution of the tuitions, fees, and room and board charges of a random sample of public 4 - year…

Explanation:

Step1: Identify mid - points and frequencies

Let's assume the mid - points and frequencies from the pie - chart information:

Class Interval	Mid - point ($x$)	Frequency ($f$)
$15000 - 17499$	$\frac{15000 + 17499}{2}=16249.5$	$8$
$17500 - 19999$	$\frac{17500+19999}{2}=18749.5$	$12$
$20000 - 22499$	$\frac{20000 + 22499}{2}=21249.5$	$17$
$22500 - 24999$	$\frac{22500+24999}{2}=23749.5$	$10$
$25000$ or more	$26249.5$	$5$

Step2: Calculate the sample mean $\bar{x}$

The formula for the sample mean of a frequency distribution is $\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i = 1}^{n}f_i}$. $\sum_{i = 1}^{n}f_ix_i=8\times16249.5 + 12\times18749.5+17\times21249.5+10\times23749.5+5\times26249.5$ $=129996+224994+361241.5+237495+131247.5$ $=1084974$ $\sum_{i = 1}^{n}f_i=8 + 12+17+10+5=52$ $\bar{x}=\frac{1084974}{52}\approx20864.88$ (already given)

Step3: Calculate the sample standard deviation $s$

The formula for the sample standard deviation of a frequency distribution is $s=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}}$, where $n=\sum_{i = 1}^{n}f_i$. $(x_1-\bar{x})^2=(16249.5 - 20864.88)^2=(- 4615.38)^2 = 21301775.94$ $(x_2-\bar{x})^2=(18749.5 - 20864.88)^2=(-2115.38)^2 = 4474896.54$ $(x_3-\bar{x})^2=(21249.5 - 20864.88)^2=(384.62)^2 = 147932.54$ $(x_4-\bar{x})^2=(23749.5 - 20864.88)^2=(2884.62)^2 = 8321092.14$ $(x_5-\bar{x})^2=(26249.5 - 20864.88)^2=(5384.62)^2 = 28994177.94$

$\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2=8\times21301775.94+12\times4474896.54+17\times147932.54+10\times8321092.14+5\times28994177.94$ $=170414207.52+53698758.48+2514853.18+83210921.4+144970889.7$ $=454897630.28$

$s=\sqrt{\frac{454897630.28}{52 - 1}}=\sqrt{\frac{454897630.28}{51}}\approx2982.77$

Answer:

$2982.77$