the distribution of the tuitions, fees, and room and board charges of a random sample of public 4 - year…
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: $15,000 - $17,499, $17,500 - $19,999, $20,000 - $22,499, $22,500 - $24,999, $25,000 or more; x: 16249.5, 18749.5, 21249.5, 23749.5, 26249.5; f: 9, 11, 18, 10, 5. the sample mean is x = $20824.97. (round to the nearest cent.) the sample standard deviation is s = $ (round to the nearest cent.)
Answer
Explanation:
Step1: Recall sample - mean formula for grouped data
The formula for the sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i = 1}^{n}f_i}$, where $f_i$ is the frequency of the $i$-th class and $x_i$ is the mid - point of the $i$-th class. We have the following data:
| Class | Mid - point ($x_i$) | Frequency ($f_i$) | $f_ix_i$ |
|---|---|---|---|
| $15000 - 17499$ | $16249.5$ | $9$ | $16249.5\times9 = 146245.5$ |
| $17500 - 19999$ | $18749.5$ | $11$ | $18749.5\times11=206244.5$ |
| $20000 - 22499$ | $21249.5$ | $18$ | $21249.5\times18 = 382491$ |
| $22500 - 24999$ | $23749.5$ | $10$ | $23749.5\times10=237495$ |
| $25000$ or more | $26249.5$ | $5$ | $26249.5\times5 = 131247.5$ |
| $\sum_{i = 1}^{n}f_i=9 + 11+18 + 10+5=53$ | |||
| $\sum_{i = 1}^{n}f_ix_i=146245.5+206244.5 + 382491+237495+131247.5=1103723.5$ | |||
| $\bar{x}=\frac{1103723.5}{53}\approx20824.97$ |
Step2: Recall sample - standard - deviation formula for grouped data
The formula for the sample standard deviation $s=\sqrt{\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}}$, where $n=\sum_{i = 1}^{n}f_i$. First, calculate $(x_i-\bar{x})^2$ and $f_i(x_i - \bar{x})^2$ for each class: For $x_1 = 16249.5$, $(x_1-\bar{x})^2=(16249.5 - 20824.97)^2=( - 4575.47)^2 = 20934917.72$ and $f_1(x_1-\bar{x})^2=9\times20934917.72 = 188414259.48$ For $x_2 = 18749.5$, $(x_2-\bar{x})^2=(18749.5 - 20824.97)^2=( - 2075.47)^2=4307551.72$ and $f_2(x_2-\bar{x})^2=11\times4307551.72 = 47383068.92$ For $x_3 = 21249.5$, $(x_3-\bar{x})^2=(21249.5 - 20824.97)^2=(424.53)^2 = 180225.72$ and $f_3(x_3-\bar{x})^2=18\times180225.72=3244062.96$ For $x_4 = 23749.5$, $(x_4-\bar{x})^2=(23749.5 - 20824.97)^2=(2924.53)^2=8553898.72$ and $f_4(x_4-\bar{x})^2=10\times8553898.72 = 85538987.2$ For $x_5 = 26249.5$, $(x_5-\bar{x})^2=(26249.5 - 20824.97)^2=(5424.53)^2=29425542.72$ and $f_5(x_5-\bar{x})^2=5\times29425542.72=147127713.6$ $\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2=188414259.48+47383068.92+3244062.96+85538987.2+147127713.6=471708092.16$ $s=\sqrt{\frac{471708092.16}{53 - 1}}=\sqrt{\frac{471708092.16}{52}}\approx3011.37$
Answer:
The sample standard deviation is $s\approx3011.37$