**Property 1**: If *x̄* is the mean of the sample *S = *{*x _{1}, x_{2}, …, x_{n}*}, then the sample variance can be expressed by

it follows that

**Property 2**: If *µ* is the mean of the population *S = *{*x _{1}, x_{2}, …, x_{n}*}, then the population variance can be expressed by

Proof: Similar to the proof of Property 1.

**Property 3**: If the population {*x _{1}, x_{2}, …, x_{n}*} has mean

*µ*and standard deviation

_{x}*σ*and the population {y

_{x}_{1}, y

_{2}, …, y

*} has mean*

_{m}*µ*

_{y }and standard deviation

*σ*

_{y}, then the variance of the combined population is

Thus if *µ _{x}* =

*µ*

_{y }the combined population variance would be

**Property 4**: If the sample {*x _{1}, x_{2}, …, x_{n}*} has mean

*x̄*and standard deviation

*s*

_{x}and the sample {y

_{1}, y

_{2}, …, y

*} has mean*

_{m}*ȳ*and standard deviation

*s*

_{y}, then the variance of the combined sample is

Thus if *x̄* = ȳ, the combined sample variance would be

Proof: In general for a sample of size *k*, the sample variance can be calculated in the same way as the population variance of the same size, except that the result needs to be multiplied by *k*/(*k* – 1). Thus from the formula in Property 3 we get

from which the result follows easily.

Dr. Zaiontz,

Thank you, this has helped me so much. One thing I’d like to know is how I can calculate the standard deviation of multiple samples with different ns, mus, and sigmas. Property 4 shows how you can do it with 2 samples and I’m working with 51, and I could apply Property 4 50 times to combine all 51 samples, but I was wondering if there was a less cumbersome manner of doing this.

Thanks,

Cave

Cave,

I have just added a new example (Example 5) to the Measures of Variability webpage, which shows how to perform the calculation you are looking for. I have calculated the variance, but the standard deviation is similar. Thanks for your question.

Charles

In my opinion the formulas for property 3 to have errors in the term:

n*var(X) + m*var(Y),

whereas the term should have been:

m*var(X) + n*var(Y),

given that x has m members and y has n members.

The same error of applying the wrong membership counts to X and Y exists in property 4 as well.

James,

You are correct. X should have n elements and Y should have m elements. I have now corrected the website. Thanks very much for catching this error.

Charles