Respuesta :
Answer:
A) Private colleges:
mean = 42.7, SD = 6.72
In thousand dollars:
mean = $42700, SD= $6720
B) Public colleges:
mean = 22.3, SD = 4.53
In thousand dollars:
mean = $22300, SD= $4530
Step-by-step explanation:
A) Private Colleges:
Mean:
Total no. of samples = n =10
Sample values in dollar = x= [53.8, 42.2, 44.0, 34.3, 44.0,31.6, 45.8, 38.8, 50.5, 42.0]
Sum of samples = ∑x= 427
[tex]sample\,\,mean = \bar{x} =\frac{\sum x}{n}\\\\\bar{x}=\frac{427}{10}\\\\\bar{x}=42.7\\[/tex]
Sample mean in thousand dollars is $ 42700.
Standard Deviation:
Formula for standard deviation of sample data is
[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_{i}-\bar{x})^2}\\\\\sum(x_i-\bar{x})^2=(53.8-42.7)^2+(42.2-42.7)^2+...+(42.0-42.7)^2\\\\\sum(x_i-\bar{x})^2=123.21+0.25+ 1.69+ 70.56+ 1.69+ 123.21+ 9.61+ 15.21+60.84+0.49\\\\\sum(x_i-\bar{x})^2=406.67\\\\\sigma=\sqrt{\frac{406.67}{9}}\\\\\sigma=6.72[/tex]
Standard deviation in thousand dollars is $ 6720.
B) Public Colleges:
Mean:
Total no. of samples = n =12
Sample values in dollar = x= [20.3, 22.0, 28.2, 15.6, 24.1, 28.5,22.8, 25.8, 18.5, 25.6, 14.4, 21.8]
Sum of samples = ∑x= 267.6
[tex]sample\,\,mean = \bar{x} =\frac{\sum x}{n}\\\\\bar{x}=\frac{267.6}{12}\\\\\bar{x}=22.3\\[/tex]
Sample mean in thousand dollars is $ 22300.
Standard Deviation:
[tex]\sigma=\sqrt{\frac{1}{N-1}\sum_{i=1}^{N}(x_{i}-\bar{x})^2}\\\\\sum(x_i-\bar{x})^2=(20.3-22.3)^2+(22.0-22.3)^2+...+(21.8-22.3)^2\\\\\sum(x_i-\bar{x})^2=4+ 0.09+34.81+ 44.89+ 3.24+ 38.44+0.25+ 12.25+14.44+10.89+62.41+0.25\\\\\sum(x_i-\bar{x})^2=225.96\\\\\sigma=\sqrt{\frac{225.96}{11}}\\\\\sigma=4.53[/tex]
Standard deviation in thousand dollars is $ 4530.