Measures of Variability Detailed

Property 1: If   is the mean of the sample S = {x1, x2, …, xn}, then the sample variance can be expressed by


Proof: Since

it follows that


Property 2: If µ is the mean of the population S = {x1, x2, …, xn}, then the population variance can be expressed by


Proof: Similar to the proof of Property 1.

Property 3: If the population {x1, x2, …, xn} has mean µx and standard deviation σx and the population {y1, y2, …, ym} has mean µand standard deviation σy, then the variance of the  combined population is


Thus if µx = µthe combined population variance would be


Property 4: If the sample {x1, x2, …, xn} has mean   and standard deviation sx and the sample {y1, y2, …, ym} has mean ȳ and standard deviation sy, then the variance of the combined sample is


Thus if  = 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.

4 Responses to Measures of Variability Detailed

  1. Cave Forwin says:

    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.


    • Charles says:

      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.

  2. James Ryer says:

    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.

    • Charles says:

      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.

Leave a Reply

Your email address will not be published. Required fields are marked *