Question 453714: I need some assistance with the following problem. My work doesn't look correct.
For the following scores find the (a) mean, (b) median, and (c) sum of the squared deviations. 3.0, 3.4, 2.6, 3.3, 3.2, 3.2
(a) mean
2.6, 3.0, 3.2, 3.3, 3.4, 3.5
Number of scores 6
Total: 19/6 M=3.17
(b) median
2.6, 3.0, 3.2, 3.3, 3.4, 3.5
6+1=7 divide by 2 = 3.5
(c) Sum of squared deviations
1st term: (2.6-3.17)= (-.57)^2= .3249
2nd term: (3.0-3.17)= (-.17)^2= .029
3rd term: (3.2-3.17)= (-.03)^2=.009
4th term: (3.3-3.17) (.13)^2=.169
5th term: (3.4-3.17)=(.23)^2=.0529
6th term:(3.5-3.17)=(.33)^2=.1089
Thank you for your help
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! For the following scores find the (a) mean, (b) median, and (c) sum of the squared deviations. 3.0, 3.4, 2.6, 3.3, 3.2, 3.2
(a) mean
2.6, 3.0, 3.2, 3.3, 3.4, 3.5
Number of scores 6
Total: 19/6 M=3.17
(b) median is the middle score.
You have 2 values in the middle.
Use the average of those 2 values.
2.6, 3.0, 3.2, 3.3, 3.4, 3.5
(3.2+3.3)/2 = 3.25
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(c) Sum of squared deviations
1st term: (2.6-3.17)= (-.57)^2= .3249
2nd term: (3.0-3.17)= (-.17)^2= .029
3rd term: (3.2-3.17)= (-.03)^2=.009
4th term: (3.3-3.17) (.13)^2=.169
5th term: (3.4-3.17)=(.23)^2=.0529
6th term:(3.5-3.17)=(.33)^2=.1089
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Those are the squares. You need the sum of
the squares.
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Cheers,
Stan H.
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