# Calculating missing data points from standard deviation and mean

I have to figure out what to missing data points are from a set of 10 scores. The mean of the 10 observed scores is 20.0 and the standard deviation is 6.0. The observed scores are listed below, with the two missing scores shown as a blanks. What formula/method do I use to determine these scores? We learned basic standard deviation calculations in class but I can't figure out how to reverse it to find these two scores.

{13, 11, 20, 24, 29, 27, 16, 20, _, _ }

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Hint: The mean is the sum of the scores divided by the number, so if the mean is 20 the total is 200. Let the remaining scores be $x$ and $y$ and this gives you $x+y$. Then the standard deviation is the square root of the variance. The variance is the sum of the squares of the difference of each square from the mean. So you have $(13-20)^2+(11-20)^2+\ldots+(x-20)^2+(y-20)^2=$what?

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Ross, you seem to have written $10$ where you meant $20$ in the variance formula. I hope you don't mind my edit to fix it. – B R Oct 1 '12 at 4:03
@B.R.: you are right. Thanks – Ross Millikan Oct 1 '12 at 4:12

Okay. this is how you do it. use this formula 1/n SIGMA xi^2-meansquared okay? okay. n= the number of numerical thingies sothe mean, dear dude, is 160+a+b/10 = 20 160+a+b = 200 a+b = 40 b= 40-a EQN 1 1/10 (13)^2+(11)^2+(20)^2+(24)^2+(27)^2+(16)^2+(20)^2+(a)^2+(b)^2- 20=6^2 1/10(2651+a^2+b^2) - 20 = 36 2651+a^2+b^2 - 200= 360 a^2+b^2+2001=0 substitute EQN 1 into this so a^2+(40-a)^2+2001=0 a^2+1600-80a+a^2+2001=0 2a^2-80a+3601=0 now solve this. and you will get two values, they can be used interchangeably

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Hi Kari, welcome to Math.SE! Your post will be easier to read if you format the math using LaTeX. Even without LaTeX, you can get linebreaks by leaving two spaces at the end of the line you want a break after. – Jonathan Christensen Jan 22 '13 at 17:56