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Math Help - Normal Distribution

  1. #1
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    Normal Distribution

    A cylinder is to be machined to a diameter of 8cm. The Upper specifications limit is 8.2cm and the lower specifications limit is 7.9cm. A certain machine produces cylinders with a mean of 8.1cm with a standard deviation of 0.1cm.

    What is the probability of being outside the tolerances for the cylinder?

    I know P(X>8.2) and that i then need to convert it to a Z score with z=X-mean/Standard Deviation = z=8.2-8.1/0.1 = 1
    I get stuck from there.

    Any help would be greatly appreciated
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  2. #2
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    Quote Originally Posted by mortalcyrax View Post

    I know P(X>8.2) and that i then need to convert it to a Z score with z=X-mean/Standard Deviation = z=8.2-8.1/0.1 = 1
    I get stuck from there.
    This sounds good, you have to use the 68%-95%-99.7% rules here or look up a table. This online calculator will help.

    Z table - Normal Distribution
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  3. #3
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    Quote Originally Posted by mortalcyrax View Post
    A cylinder is to be machined to a diameter of 8cm. The Upper specifications limit is 8.2cm and the lower specifications limit is 7.9cm. A certain machine produces cylinders with a mean of 8.1cm with a standard deviation of 0.1cm.

    What is the probability of being outside the tolerances for the cylinder?

    I know P(X>8.2) and that i then need to convert it to a Z score with z=X-mean/Standard Deviation = z=8.2-8.1/0.1 = 1
    I get stuck from there.

    Any help would be greatly appreciated
    Maybe your z-tables only deal with positive values of z.

    The probability of being outside the tolerances is P(z<-2)+P(z>1), since \frac{7.9-8.1}{0.1}=-2

    The values given by the z-tables are P(z<value) or P(z\ \le\ value)
    which are basically the same as the curve is continuous.

    However P(z<-2)=P(z>2) due to the symmetry of the bell-shaped curve.

    and P(z>2)=1-P(z<2)

    Also P(z>1)=1-P(z<1)

    Hence the probability of being machined outside of specifications is

    1-P(z<2)+1-P(z<1)=2-P(z<2)-P(z<1)
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  4. #4
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    So P(X>8.2)= P(Z>1)

    = 0.5 - P(Z>1)

    = 0.5 - 0.3413

    = 0.1587 <--- probability of being machined above the specifications

     P(X<7.9)=P(Z< -2)

    = 0.5 - 0.4772

    = 0.0228 <--- probability of being machined lower than the specifications

    Then add the two probabilities to get the probability of being machined outside of the specifications.

    0.1587 + 0.0228 = 0.1815
    Is this right???
    Last edited by mortalcyrax; August 3rd 2010 at 11:41 PM. Reason: Found Mistakes
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  5. #5
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    Yes,
    that's what I get also.

    You're calculating the probability of z being between 0 and 1
    then subtracting from 0.5 to get the probability of being above 1....
    and z being between 0 and -2
    and subtracting from 0.5 to find the probability of z being below -2.

    Excellent.
    Last edited by Archie Meade; August 4th 2010 at 06:54 AM.
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