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.)\nclass\n$15,000 - $17,499\n$17,500 - $19,999\n$20,000 - $22,499\n$22,500 - $24,999\n$25,000 or more\nx\nf

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.)\nclass\n$15,000 - $17,499\n$17,500 - $19,999\n$20,000 - $22,499\n$22,500 - $24,999\n$25,000 or more\nx\nf

Answer

Explanation:

Step1: Calculate mid - points

For the class $15000 - 17499$, the mid - point $x_1=\frac{15000 + 17499}{2}=16249.5$. For the class $17500 - 19999$, the mid - point $x_2=\frac{17500+19999}{2}=18749.5$. For the class $20000 - 22499$, the mid - point $x_3=\frac{20000 + 22499}{2}=21249.5$. For the class $22500 - 24999$, the mid - point $x_4=\frac{22500+24999}{2}=23749.5$. For the class $25000$ or more, we are given the mid - point $x_5 = 26249.5$.

Step2: Calculate sample mean

Let the frequencies be $f_1 = 11,f_2=10,f_3 = 18,f_4=9,f_5 = 7$. The formula for the sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}f_ix_i}{\sum_{i=1}^{n}f_i}$. $\sum_{i = 1}^{5}f_ix_i=f_1x_1+f_2x_2+f_3x_3+f_4x_4+f_5x_5$ $=11\times16249.5+10\times18749.5+18\times21249.5+9\times23749.5+7\times26249.5$ $=178744.5+187495+382491+213745.5+183746.5$ $=1146222.5$. $\sum_{i=1}^{5}f_i=11 + 10+18+9+7=55$. $\bar{x}=\frac{1146222.5}{55}\approx20840.41$.

Step3: Calculate sample variance

The formula for the sample variance $s^2=\frac{\sum_{i = 1}^{n}f_i(x_i-\bar{x})^2}{n - 1}$, where $n=\sum_{i=1}^{n}f_i$. $(x_1-\bar{x})=16249.5 - 20840.41=-4590.91$, $(x_2-\bar{x})=18749.5 - 20840.41=-2090.91$, $(x_3-\bar{x})=21249.5 - 20840.41 = 409.09$, $(x_4-\bar{x})=23749.5 - 20840.41=2909.09$, $(x_5-\bar{x})=26249.5 - 20840.41 = 5409.09$. $f_1(x_1-\bar{x})^2=11\times(-4590.91)^2=11\times21076477.13=231841248.43$. $f_2(x_2-\bar{x})^2=10\times(-2090.91)^2=10\times4371833.63=43718336.3$. $f_3(x_3-\bar{x})^2=18\times(409.09)^2=18\times167354.62=3012383.16$. $f_4(x_4-\bar{x})^2=9\times(2909.09)^2=9\times8462751.73=76164765.57$. $f_5(x_5-\bar{x})^2=7\times(5409.09)^2=7\times29268229.73=204877608.11$. $\sum_{i = 1}^{5}f_i(x_i-\bar{x})^2=231841248.43+43718336.3+3012383.16+76164765.57+204877608.11=559614341.57$. $s^2=\frac{559614341.57}{54}\approx10363228.55$. The sample standard deviation $s=\sqrt{s^2}=\sqrt{10363228.55}\approx3219.2$.

Answer:

Sample mean: $$20840.41$, Sample standard deviation: $$3219.2$