# variance and standard deviation

• Aug 7th 2007, 08:22 AM
0123
variance and standard deviation
Why do we average by divivding by n-1 rather than n in calculating the variance? the formula I am using is s^2=[ sum of (x- X)^2]/(n-1) where X= mean. Thank you very much for your help!
• Aug 7th 2007, 09:07 AM
CaptainBlack
Quote:

Originally Posted by 0123
Why do we average by divivding by n-1 rather than n in calculating the variance? the formula I am using is s^2=[ sum of (x- X)^2]/(n-1) where X= mean. Thank you very much for your help!

Because you are using s^2 to estimate the population variance, and dividing
by n results in a biased estimate while dividing by n-1 gives an unbiased
estimator.

RonL
• Aug 7th 2007, 12:21 PM
0123
could you be more clear? because I do not understand why it is unbiased in the case on n-1. Thanks again.
• Aug 7th 2007, 01:03 PM
CaptainBlack
Quote:

Originally Posted by 0123
could you be more clear? because I do not understand why it is unbiased in the case on n-1. Thanks again.

Take my word for it, the proof is longer than I am prepared to reproduce here.

I can illustrate it by example:

Code:

```>x=normal(100,5); >m=sum(x)/5;        ..100 means of samples of size 5 >v=sum(x^2)/5-m^2;  ..100 (1/n) variances of samples of size 5 >vv=sum(s')/100      .. average of 100 (1/n) variance of sample of size 5         0.80403236 >vv*5/4              .. corrected to 1/(n-1) variances         1.00504045 >```
Here we see the average variance of samples of size 5 from a population of
variance 1 is ~0.8, while the corrected average variance is ~1, as required.

RonL