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    This is question 7 from the May 2015 IAL S3 paper.

    A fair six-sided die is labelled with the numbers 1, 2, 3, 4, 5 and 6

    The die is rolled 40 times and the score, S, for each roll is recorded.

    (a) Find the mean and the variance of S.(2)

    (b) Find an approximation for the probability that the mean of the 40 scores is less than 3(3)

    Part a is, embarrassingly, the part I am struggling with. I use the standard formula for variance of a uniform distribution

    \frac{(b-a)^2}{12}

    To get

    \frac{(6-1)^2}{12} = \frac{25}{12}

    however the mark scheme has the answer as

    \frac{35}{12}

    Not sure if I am being a complete idiot but this looks like it should be fairly simple. Thanks for any input.
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    The formula that you have used is for a continuous uniform distribution, unfortunately this is discrete. Try using E[x^2] - (E[x])^2 and the answer should fall out. Hope this helps.
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    (Original post by BvV98)
    The formula that you have used is for a continuous uniform distribution, unfortunately this is discrete. Try using E[x^2] - (E[x])^2 and the answer should fall out. Hope this helps.
    Of course! Knew it would be a silly mistake, thanks for pointing this out
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    (Original post by Mav1)
    This is question 7 from the May 2015 IAL S3 paper.

    A fair six-sided die is labelled with the numbers 1, 2, 3, 4, 5 and 6

    The die is rolled 40 times and the score, S, for each roll is recorded.

    (a) Find the mean and the variance of S.(2)

    (b) Find an approximation for the probability that the mean of the 40 scores is less than 3(3)

    Part a is, embarrassingly, the part I am struggling with. I use the standard formula for variance of a uniform distribution

    \frac{(b-a)^2}{12}

    To get

    \frac{(6-1)^2}{12} = \frac{25}{12}

    however the mark scheme has the answer as

    \frac{35}{12}

    Not sure if I am being a complete idiot but this looks like it should be fairly simple. Thanks for any input.

    For a discrete variable the variance is:

    \displaystyle Var(X) = \sum^n_{i=1}p_i(x_i - \mu)^2

    where \mu is the expectation value. You'll find that it gives the correct result.

    (for a continuous variable it is analagous - as each individual probability of an outcome tends towards zero, the sum tends towards an integral)
 
 
 
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