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Combined Standard Deviation

  1. Jul 22, 2010 #1
    Hi, I'm having a little trouble with this question:

    Among entering students at a certain college, the men averaged 650 on the Math SAT, and their SD was 125. The women averaged 600, but had the same SD of 125. There were 500 men in he class, and 500 women.

    Q) For the men and the women together, was the SD of Math SAT scores less than 125, just about 125, or more than 125?

    I'm not quite sure how to solve this. By looking at the question, I thought the SD would stay the same. Then I pictured a normal curve for men, women, and combined, and since the combined average is 625, the SD seemed to be larger because the combined curve had to take care of the extremes of both men and women. Then out of desperation, I tried finding the pooled Standard Deviation and found it to be 125.

    I'm just going around in circles, not knowing exactly what I'm doing. I think it's either more than 125 or equal to 125, but even if one of those were right, I'm not confident on the reasoning behind it. Can you please help me solve this question, and future questions like these?
  2. jcsd
  3. Jul 22, 2010 #2


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    Well, looking at the question isn't enough. SDis a tricky thing, it doesn't behave as nicely as averages (e.g. you are allowed to say the combined average is (600 + 650)/2 = 625, but not that the combined SD is (125 + 125)/2 = 125, for instance.

    That's better. If you consider the SD as telling you something about the spread in answers, then indeed it should be larger since statistically speaking, values will be further away from the average of 625.

    That's because, as I said, you cannot simply take the average of a standard deviation. If you add two independent* variables X and Y, then for the average of the sum, E(X + Y) = E(X) + E(Y). Taking the average you get E(X + Y) / 2 = (E(X) + E(Y)) / 2.
    For the standard deviation of X + Y, there is the formula
    σ(X + Y)² = σ(X)² + σ(Y)²,
    so you don't simply add them, but you add the squares (and then take the square root).
    That gives you, in this case where σ(X) = σ(Y),
    [tex]\sigma(X + Y) = \sqrt{2 \sigma(X)^2} = \sqrt{2} \sigma(X)[/tex]
    so the standard deviation of the total score increases by a factor sqrt(2). This corresponds to your intuition, hopefully, that when adding two things with uncertainty, the uncertainty of the sum will exceed the uncertainty of a single variable - the counter-intuitive catch is that it doesn't increase by a factor of 2 but only its square root.
    For the SD of the average score, then, you divide by 2 to get
    [tex]\sigma = \frac{\sqrt{2} \sigma(X)}{2} = \frac{\sigma(X)}{\sqrt{2}}[/tex]

    Actually these laws, in the case of adding or averaging n variables with the same distribution as X read
    [tex]E(\text{sum of }n) = n E(X), \sigma(\text{sum of }n) = \sqrt{n} \sigma(X)[/tex];
    [tex]E(\text{average of }n) = E(X), \sigma(\text{average of }n) = \frac{1}{\sqrt{n}} \sigma(X)[/tex];
    and are commonly referred to as [itex]\sqrt{n}[/itex]-law.

    * Strictly speaking, E(X + Y) = E(X) + E(Y) also holds when X and Y are not independent, however for the SD independence is necessary.
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