A1 The following data values are the ages and gender MF of t

A1. The following data values are the ages and gender (M/F) of the students in one section of STAT 1490 from a previous semester.

25M, 20F, 23F, 21F, 18F, 20M, 21M, 21M, 25F, 19M, 23M, 20F, 31F, 21F, 19M, 18F, 19F, 22M, 19F, 18M, 19F, 19F, 20F, 23M, 21F, 19F, 19F

Find the mean, variance, and standard deviation for these data

Solution

  
Setting up tables,          
x   x - X   (x - X)^2  
25   4.148148148   17.20713306  
20   -0.851851852   0.725651578  
23   2.148148148   4.614540466  
21   0.148148148   0.021947874  
18   -2.851851852   8.133058985  
20   -0.851851852   0.725651578  
21   0.148148148   0.021947874  
21   0.148148148   0.021947874  
25   4.148148148   17.20713306  
19   -1.851851852   3.429355281  
23   2.148148148   4.614540466  
20   -0.851851852   0.725651578  
31   10.14814815   102.9849108  
21   0.148148148   0.021947874  
19   -1.851851852   3.429355281  
18   -2.851851852   8.133058985  
19   -1.851851852   3.429355281  
22   1.148148148   1.31824417  
19   -1.851851852   3.429355281  
18   -2.851851852   8.133058985  
19   -1.851851852   3.429355281  
19   -1.851851852   3.429355281  
20   -0.851851852   0.725651578  
23   2.148148148   4.614540466  
21   0.148148148   0.021947874  
19   -1.851851852   3.429355281  
19   -1.851851852   3.429355281  

Getting the mean, X,          
          
X = Sum(x) / n          
Sum(x) =    563      
          
Thus,          
X =    20.85185185 [answer, mean]
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Thus, Sum(x - X)^2 =    207.4074074      
          
Thus, as           
          
s^2 = Sum(x - X)^2 / (n - 1)          
          
As n =    27      
          
s^2 =    7.977207977 [variance]
          
Thus,          
          
s =    2.824395152 [standard deviation]