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Math Help - Another qn!

  1. #1
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    Another qn!

    Jack drives down to town every morning. On 16 randomly chosen trips, he recorded his travel time xi (i is a subscript!) in minutes.

    He finds that \displaystyle\sum_{i=1}^{16} xi = 519 and \displaystyle\sum_{i=1}^{16} (xi)^2 = 16983

    Calculate the unbiased estimates of the variance of the driving times to town.
    I could calculate the unbiased estimate of the mean, but I dont get how to get the variance from \displaystyle\sum_{i=1}^{16} (xi)^2 = 16983 . What can I get from the sum of (xi)^2? I know I can easily get the mean from the sum of xi, but (xi)^2? Isn't the variance the sum of ((xi-mean)^2) / n so it has nothing to do with the sum of (xi)^2?

    Thank you so much in advance, really appreciate this!
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  2. #2
    GJA
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    Re: Another qn!

    Hi Tutu,

    I think you know how to answer your own question, so I'm going to do my best to give you a hint without telling you what to do. You said,

    Isn't the variance the sum of ((xi-mean)^2) / n so it has nothing to do with the sum of (xi)^2?
    But \frac{1}{n}\sum(x_{i}-\mu)^{2} does have something to do with \sum x_{i}^{2}. Stare at (x_{i}-\mu)^{2} for a minute and I think you'll see what needs to happen.

    Let me know if anything is unclear. Good luck!
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  3. #3
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    Re: Another qn!

    Hi I stared. I can infer that I am to subtract the mean from every xi value. And I know the mean already. But how to, if they already give me the sum of (xi)^2? I am unsure about how to insert the equation..Is it (xi-Mean)^2 = 16983?
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  4. #4
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    Re: Another qn!

    I looked at a few videos that manipulated the variance formula. Decided to use (sum of (xi)^2)/N - mean^2 as the variance. However, I got 11.7, which is still not the answer..
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  5. #5
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    Re: Another qn!

    Apparently this formula will get me the answer, 9.84.
    (N * sum of all x - (sum of all x's)) / (N * (N-1))

    Can someone tell me how this formula is the variance? How is it derived algebraically from the variance? Please help me, really.

    Thank you much in adva
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