Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Statistics - normal distribution

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data
    A plant manufactures 500 components a day with the diameter being random variable:
    N(8.02, 0.1^2)mm

    What is the probability of two randomly picked components differing by more than 0.3mm?

    2. Solution
    I know that the solution is 0.966

    3. The attempt at a solution
    I thought that I needed to find the probability of the diameter differing from the mean by more than 0.3mm:
    P(X < 7.72) + P(X > 8.32)
    = 1 - P(7.72 < X < 8.32)
    = 1 - { Phi[(8.32-8.02)/0.1] - Phi[(7.72-8.02)/0.1] }
    = 1 - { Phi[3] - Phi[-3] }
    = 1 - { Phi[3] - (1 - Phi[3]) }
    = 1 - {0.99865 - (1-0.99865)}
    = 1 - 0.9973
    = 0.0027

    This obviously is far from the correct answer and so I realise I must be completely on the wrong track but don't know how else one might approach this problem. I thought that the question might instead be asking for the probability of the difference of the two diameters being greater than 0.3mm so
    P(a < X < a+0.3) but have no idea how you would go about finding the probability of an unknown value.

    As you can probably tell, I am really confused by this so any help would be much appreciated! Thank you!
  2. jcsd
  3. Mar 23, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    You have two random variables: X1, the diameter of the first ball, and X2, the diameter of the second ball. Now form a third random variable Y=X1-X2, which is the difference in their diameters. Do you see how to solve the problem now?
  4. Mar 23, 2010 #3
    I think I might have got it now...

    So for this new variable Y=X1-X2, I have to also calculate a new standard deviation which is sqrt(0.12 + 0.12) = 0.1*sqrt(2)

    So P(X>0.3) = 1 - P(X<0.3)
    = 1 - Phi[0.3/(0.1*sqrt(2))]
    = 1 - Phi[2.1213...]
    = 1 - 0.983
    = 0.0169
    P(X<-0.3) = 0.0169
    P(-0.3<X<0.3) = 1 - 2*0.0169
    = 0.9661 (I guess the solution must have just been given for the difference being less than 0.3mm??)

    Is this right? I understand why this would be correct but then when I apply this to the next part of the question, I do not get the right answer.
    Now another plant B produces components with diameter being random variable N(7.95,0.082)

    What now is the probability of the diameter of two randomly picked components differing by more than 0.3mm if one is produced by plant B and the other by the first plant?

    I went about this in the same way, calculating a standard deviation for this distribution:
    sqrt(0.12 + 0.082) = [sqrt(41)]/50

    Using the mean as zero and this standard deviation, I get P(-0.3<X<0.3)=0.9808, whereas the solution I am given is 0.9662.....where am I going wrong

    Sorry for the long post!
  5. Mar 23, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    This is where you're going wrong:

  6. Mar 23, 2010 #5
    Ahhhhhh THANK YOU SO, SO MUCH! I have been struggling with this question for so long!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook