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  1. #1
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    Unhappy Please help

    Raisins are packaged into cardboard containers. The weight of raisins delivered into a container has a distribution with mean 200g (grams) and a standard deviation of 6g. The weight of an empty container has a distribution with mean 15g and a standard deviation of 0.5g. Assume the weight of the empty container and the weight of the raisins are independent.

    Calculate..

    a) the expected weight of a filled container of raisins.

    b) the variance of a filled contained of raisins.

    c) the standard deviation of a filled container of raisins.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by norivea
    Raisins are packaged into cardboard containers. The weight of raisins delivered into a container has a distribution with mean 200g (grams) and a standard deviation of 6g. The weight of an empty container has a distribution with mean 15g and a standard deviation of 0.5g. Assume the weight of the empty container and the weight of the raisins are independent.

    Calculate..

    a) the expected weight of a filled container of raisins.
    The expected weight is the same thing as the mean weight.
    The mean of the sum of two random variables is the sum of their means
    so in this case it is 200+15=215\mathrm{g}.

    b) the variance of a filled contained of raisins.
    The variance of the sum of two independent random variables is the sum
    of the variances. The variance is the square of the standard deviation so
    in this case the required variance is 6^2+0.5^2=36.25\mathrm{g^2}

    c) the standard deviation of a filled container of raisins.
    This is the square root of the variance so is \sqrt{36.25}\approx6.021\mathrm{g}

    RonL
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