# Needing to find a variance from a data set below

I am trying to get the variance of the data set below. I get a negative number upon applying 1/2(2.5)^2 + 1(3)^2 + 0.5(4.5)^2 + 1(5)^2 + 1(6)^2 - (17.5)^2 = variance, but a negative and weird number.

Please use this data below from the link, the problem is as is and the sample space is what I also got. But I got a mean of 17.5 and not sure how to to do the variance.

http://answers.yahoo.com/question/index?qid=20100120103118AAld5Yx

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## 1 Answer

The problem is to consider all possible samples of size 2 from the set of values 1,2,2,4,8 and find the distribution, mean and variance of the number of values >3. The set of possible samples are: [1,2] [1.2], [1.4], [1,8], [2,2], [2,4], [2,8]. [2,4]. [2,8] amd [4,8]. These 10 samples are equally likely. The values for the random variable X = # of cases > 3 are 0,0,1,1,0,1,1,1,1,2. The mean is 0.8 and the variance is [3(0-0.8)$^2$+5(1-0.8)$^2$+(2-0.8)$^2$]/10 =[3(0.64)+5(0.04)+1.44]/10= [1.92+0.20+1.44]/10= 3.56/10 =0.356.

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I got the mean to be: 1/2(2.5) + 1(3) + 1/2(4.5) + 1(5) + 1(6) = 17.5 –  mary Sep 4 '12 at 1:43
and the variance was a negative number, which I was sure that something was wrong –  mary Sep 4 '12 at 1:44
Based on the link the distribution is as I have given adn the mean is 0.8 not 17.5 and the variance is +0.356 which is certainly not negative. –  Michael Chernick Sep 4 '12 at 2:46
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