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Math Help - Really simple question too hard for a simple mind

  1. #1
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    Really simple question too hard for a simple mind

    Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept coins weighing between 5.48 g and 5.82 g. What percentage of legal quarters will be rejected?

    The answer is:
    The standard score ("Z") for 5.48:
    Z = (5.48 -5.67)/0.070 = -2.71
    The area between the -2.71 and the mean (from a standard normal table): 0.4966

    The standard score for 5.82:
    Z = (5.82-5.67)/0.070 = 2.14
    The area between 2.14 and the mean: 0.4838

    The area between 5.48 (Z=-2.71) and 5.82 (Z=2.14) = 0.4966+0.4838 = 0.9804.
    The area outside that area is 1 = 0.9804 = 0.0196.
    Therefore, about 2% of legal quarters will be rejected.



    My question is, the table for the standard normal curve areas i have is not giving me that -2.71= .4966 & 2.14=.4838.
    For me my table says (-2.71<z<2.14) = (z<2.14)=.9836 and [1-(z<2.71)]=1-.9966=.0034
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  2. #2
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    Quote Originally Posted by Aquameatwad View Post
    Assume that the weights of quarters are normally distributed with a mean of 5.67 g and a standard deviation 0.070 g. A vending machine will only accept coins weighing between 5.48 g and 5.82 g. What percentage of legal quarters will be rejected?

    The answer is:
    The standard score ("Z") for 5.48:
    Z = (5.48 -5.67)/0.070 = -2.71
    The area between the -2.71 and the mean (from a standard normal table): 0.4966

    The standard score for 5.82:
    Z = (5.82-5.67)/0.070 = 2.14
    The area between 2.14 and the mean: 0.4838

    The area between 5.48 (Z=-2.71) and 5.82 (Z=2.14) = 0.4966+0.4838 = 0.9804.
    The area outside that area is 1 = 0.9804 = 0.0196.
    Therefore, about 2% of legal quarters will be rejected.



    My question is, the table for the standard normal curve areas i have is not giving me that -2.71= .4966 & 2.14=.4838.
    For me my table says (-2.71<z<2.14) = (z<2.14)=.9836 and [1-(z<2.71)]=1-.9966=.0034
    Look at the problem like this:

    Coins that are less than 5.48g will be rejected so

    pr(x<\frac{5.48-5.67}{0.070})=pr(z<-2.71)

    Coins that are more than 5.82g will be rejected so:

    pr(x>\frac{5.82-5.67}{0.070})=pr(z>2.14)=pr(z<-2.14)

    So the rejected region is

    pr(z<-2.14)+pr(z<-2.71)

    Now use your z-table and see if you get the correct values
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    This makes much more sense. thank you very much.
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