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.\ $25,000 or more 26249.5 5 the sample mean is $\bar{x} = $20864.88$. (round to the nearest cent.) the sample standard deviation is $s = $2669.57$. (round to the nearest cent.)

Explanation:

Step1: Determine mid - points and frequencies

Class Interval	Mid - point ($x_i$)	Frequency ($f_i$)
$17500 - 19999$	$\frac{17500+19999}{2}=18749.5$	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 $\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i = 1}^{n}f_i}$.
$\sum_{i = 1}^{n}f_i=8 + 12+17+10+5=52$.
$\sum_{i = 1}^{n}f_ix_i=8\times16249.5+12\times18749.5+17\times21249.5+10\times23749.5+5\times26249.5$
$=129996+224994+361241.5+237495+131247.5$
$=1084974$.
$\bar{x}=\frac{1084974}{52}\approx20864.88$.

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 $n=\sum_{i = 1}^{n}f_i$.
First, calculate $(x_i-\bar{x})^2$ and $f_i(x_i - \bar{x})^2$ for each class:
For $x_1 = 16249.5$, $(x_1-\bar{x})^2=(16249.5 - 20864.88)^2=(- 4615.38)^2 = 21301775.0444$, $f_1(x_1-\bar{x})^2=8\times21301775.0444 = 170414200.3552$.
For $x_2 = 18749.5$, $(x_2-\bar{x})^2=(18749.5 - 20864.88)^2=(-2115.88)^2 = 4476998.1744$, $f_2(x_2-\bar{x})^2=12\times4476998.1744 = 53723978.0928$.
For $x_3 = 21249.5$, $(x_3-\bar{x})^2=(21249.5 - 20864.88)^2=(384.62)^2 = 147932.5444$, $f_3(x_3-\bar{x})^2=17\times147932.5444 = 2514853.2548$.
For $x_4 = 23749.5$, $(x_4-\bar{x})^2=(23749.5 - 20864.88)^2=(2884.62)^2 = 8321090.1444$, $f_4(x_4-\bar{x})^2=10\times8321090.1444 = 83210901.444$.
For $x_5 = 26249.5$, $(x_5-\bar{x})^2=(26249.5 - 20864.88)^2=(5384.62)^2 = 28994174.1444$, $f_5(x_5-\bar{x})^2=5\times28994174.1444 = 144970870.722$.
$\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2=170414200.3552+53723978.0928+2514853.2548+83210901.444+144970870.722$
$=454834803.8688$.
$s=\sqrt{\frac{454834803.8688}{51}}\approx2669.57$.

Answer:

The sample mean is $\$20864.88$ and the sample standard deviation is $\$2669.57$.