SOLUTION: I have a statistics problem where I am suppose to find the mean, median, and sum of squared deviation for the following scores: 8, -5,7, -10, 5. I found the mean by adding up all t

Algebra ->  Probability-and-statistics -> SOLUTION: I have a statistics problem where I am suppose to find the mean, median, and sum of squared deviation for the following scores: 8, -5,7, -10, 5. I found the mean by adding up all t      Log On


   



Question 453297: I have a statistics problem where I am suppose to find the mean, median, and sum of squared deviation for the following scores: 8, -5,7, -10, 5. I found the mean by adding up all the scores and then dividing by the numbers of scores. So the mean is M=1. My median was found by adding 5 + 1 and then divided by 2= 5.5. Finding the sum of squared deviations I'm at a lose. I know my formula is SS/N .I know N represents the total number of scores which is 5 but I'm not sure what SS is. Any assistance would be appreciated.
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
I have a statistics problem where I am supposed to find the mean, median, and sum of squared deviation for the following scores: 8, -5,7, -10, 5.
----------------------------------
Ranked List of DATA: -10,,-5,,5,,7,,8
-----
I found the mean by adding up all the scores and then dividing by the numbers of scores. So the mean is (20-15)/2 = 5/2 = 2.5
--------------
The median is the middle score: median = 5
---------------
Finding the sum of squared deviations I'm at a lose. I know my formula is SS/N .I know N represents the total number of scores which is 5 but I'm not sure what SS is. Any assistance would be appreciated.
1st term: (-10-2.5)^2 = (-12.5)^2 = 156.25
2nd term: (-5-2.5)^2 = (-7.5)^2 = 56.25
3rd term: (5-2.5)^2 = (2.5)^2 = 6.25
4th term: (7-2.5)^2 = (4.5)^2 = 20.25
5th term: (8-2.5)^2 = (5.5)^2 = 30.25
----------------------------------------
SS = the sum of those squares = 269.25
Divide by N = 5 = 53.85
==================================
Cheers,
Stan H.