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    When a butcher takes an order for a Christmas Turkey, he asks the customer what weight in kilograms the bird should be. He then sends his order to a farmer who supplies the bird of about the weight requested. For any particular weight of bird ordered, the error in kilograms may be taken to be normally distributed with mean 0kg and standard deviation 0.75kg

    Mrs Jones orders a 10kg turkey from the butcher. Find the probability that the one she gets is within 0.5kg of the weight she actually ordered.

    Here is what I attempted to do:

    X~N(0, 0.75^2)
    9.5 < x <10.5

    P(X<10.5) = P(Z<0.666) = 0.7473
    I got P(Z<0.666) from doing z = (0.5 - 0) / 0.75

    P(X>9.5) = 1 - P(X<=9.5) = 1 - P(Z<0.666) = 0.2527

    When I multiply these, it gives me 0.188 but the answer in the book is 0.495.
    Where have I gone wrong here?
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    (Original post by IDontKnowReally)
    When a butcher takes an order for a Christmas Turkey, he asks the customer what weight in kilograms the bird should be. He then sends his order to a farmer who supplies the bird of about the weight requested. For any particular weight of bird ordered, the error in kilograms may be taken to be normally distributed with mean 0kg and standard deviation 0.75kg

    Mrs Jones orders a 10kg turkey from the butcher. Find the probability that the one she gets is within 0.5kg of the weight she actually ordered.

    Here is what I attempted to do:

    X~N(0, 0.75^2)
    9.5 < x <10.5

    P(X<10.5) = P(Z<0.666) = 0.7473
    I got P(Z<0.666) from doing z = (0.5 - 0) / 0.75

    P(X>9.5) = 1 - P(X<=9.5) = 1 - P(Z<0.666) = 0.2527

    When I multiply these, it gives me 0.188 but the answer in the book is 0.495.
    Where have I gone wrong here?
    You are almost there but you do not need to multiply them. P(X&lt;9.5)=0.2527=P(X&gt;10.5). Both of these probability statements are areas under the normal distribution. The total area under the curve is 1 so the probability is 1-(area we dont want) so
    P(9.5\leq x \leq 10.5)=1-P(X&lt;9.5)-P(X&gt;10.5)=0.4946
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    (Original post by Cryptokyo)
    You are almost there but you do not need to multiply them. P(X&lt;9.5)=0.2527=P(X&gt;10.5). Both of these probability statements are areas under the normal distribution. The total area under the curve is 1 so the probability is 1-(area we dont want) so
    P(9.5\leq x \leq 10.5)=1-P(X&lt;9.5)-P(X&gt;10.5)=0.4946
    That makes a lot more sense, thank you!
    With these sorts of questions, do you always have to find what the probability is of the values it cant be?
    Or is there are a way to do it if you know that:
    P(X<0.5) = P(X>-0.5) = 0.7473
    because adding it gives an answer greater than 1?
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    (Original post by IDontKnowReally)
    That makes a lot more sense, thank you!
    With these sorts of questions, do you always have to find what the probability is of the values it cant be?
    Or is there are a way to do it if you know that:
    P(X<0.5) = P(X>-0.5) = 0.7473
    because adding it gives an answer greater than 1?
    Sorry for the late reply. It can work either way except you must be careful as in this case P(X&lt;0.5)=P(z&lt;0.666) includes area we don't want in this case (see picture). Note \Phi(z) is the normal distribution function.
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