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.

complete the frequency distribution for the data.
(type integers or decimals. do not round.)
class\t x\t f
$15,000 - $17,499\t 16249.5\t 9
$17,500 - $19,999\t 18749.5\t 11
$20,000 - $22,499\t 21249.5\t 18
$22,500 - $24,999\t 23749.5\t 10
$25,000 or more.\t 26249.5\t 5

the sample mean is x = $20824.97.
(round to the nearest cent.)

the sample standard deviation is s = $ .
(round to the nearest cent.)

Explanation:

Step1: Recall sample - standard - deviation formula

The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}}$, where $n=\sum_{i=1}^{n}f_i$, $f_i$ is the frequency of the $i$-th class, $x_i$ is the mid - point of the $i$-th class, and $\bar{x}$ is the sample mean. First, calculate $n$:
$n=9 + 11+18+10+5=53$.

Step2: Calculate $(x_i-\bar{x})^2f_i$ for each class

For the first class ($x_1 = 16249.5$, $f_1=9$, $\bar{x}=20824.97$):
$(x_1-\bar{x})^2f_1=(16249.5 - 20824.97)^2\times9=( - 4575.47)^2\times9=20934947.72\times9 = 188414529.48$.
For the second class ($x_2 = 18749.5$, $f_2 = 11$):
$(x_2-\bar{x})^2f_2=(18749.5 - 20824.97)^2\times11=( - 2075.47)^2\times11 = 4307591.72\times11=47383508.92$.
For the third class ($x_3 = 21249.5$, $f_3 = 18$):
$(x_3-\bar{x})^2f_3=(21249.5 - 20824.97)^2\times18=(424.53)^2\times18 = 180225.72\times18=3244062.96$.
For the fourth class ($x_4 = 23749.5$, $f_4 = 10$):
$(x_4-\bar{x})^2f_4=(23749.5 - 20824.97)^2\times10=(2924.53)^2\times10=8552877.72\times10 = 85528777.2$.
For the fifth class ($x_5 = 26249.5$, $f_5 = 5$):
$(x_5-\bar{x})^2f_5=(26249.5 - 20824.97)^2\times5=(5424.53)^2\times5=29425588.72\times5=147127943.6$.

Step3: Calculate the sum $\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2$

$\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2=188414529.48+47383508.92+3244062.96+85528777.2+147127943.6=471698822.16$.

Step4: Calculate the sample standard deviation

$s=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}}=\sqrt{\frac{471698822.16}{53 - 1}}=\sqrt{\frac{471698822.16}{52}}\approx\sqrt{9071131.195}\approx3011.83$.

Answer:

$3011.83$