Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

1 Answer

up vote 2 down vote accepted

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.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.