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Normal distribution problem

  1. Oct 1, 2011 #1
    1. The problem statement, all variables and given/known data

    A certain industrial process yields a large number of steel cylinders whose lengths are approximately normally distributed with a mean of 3.25 in. and a variance of 0.008 in2. If two cylinders are chosen at random and placed end to end, what is the probability that their combined length is less than 6.55 in.?


    2. Relevant equations

    normal distribution: [tex] \frac{1}{ \sigma \sqrt{2 \pi} e^{ \frac{(x- \mu )^2}{2 \sigma ^2}} }
    [/tex]


    CDF: integrate

    [itex] \mu = [/itex] mean
    [itex] \sigma = [/itex] standard deviation
    [itex] \sigma ^2 = [/itex] variance

    3. The attempt at a solution

    I'm not sure if a variance of 0.008 in2 implies [itex] \sigma = \sqrt{.008} [/itex] or [itex] \sigma = .008 [/itex].

    I also don't know how to set up the problem since there are two cylinders whose length together is less than 6.55. Do I just divide 6.55 in half? :confused:
     
    Last edited: Oct 1, 2011
  2. jcsd
  3. Oct 1, 2011 #2

    vela

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    As you noted, the variance is [itex]\sigma^2[/itex], so a variance of 0.008 in2 means [itex]\sigma^2 = 0.008~\mathrm{in}^2[/itex].
    The length of each cylinder — let's call them x1 and x2 — is a random variable with the given distribution. Their combined length z=x1+x2 is a new random variable with a related but different distribution. The problem is now asking you to find P(z ≤ 6.55 in).
     
  4. Oct 1, 2011 #3
    So I normalize by calculating [tex] P(0 \le x \le 6.55) = P( \frac{0-3.25}{.008} \le Z \le \frac{6.55-3.25}{.008} = P( -406.25 \le Z \le 412.5) [/tex]

    But my Z tables only go up to 4, which makes me feel like I'm doing something wrong. Am I?
     
  5. Oct 1, 2011 #4

    I like Serena

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    Hi Arcana! :smile:

    "z" may be an unlucky choice for the name of the variable.
    Your Z tables are tables for the "standard" normal distribution (mean zero and standard deviation of 1).
    Typically you first calculate [itex]z = {x - \mu \over \sigma}[/itex] and that is what you look up in your tables.

    But before you do that, you need to think up what the distribution is of X1 + X2, since you are taking 2 cylinders.
    Do you know how to add 2 independent normal distributions?
     
  6. Oct 1, 2011 #5
    Not a clue!

    And didn't I "standardize" the z by the calculation I made, so I could use the z table? I thought that was what that step was for.
     
  7. Oct 1, 2011 #6

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    Aah yes, you did standardize the z. My mistake. :uhh:
    You just used the wrong distribution.
    Oh, and if you "standardize" you need to divide by sigma, not by the variance.

    If you have 2 cylinders that each has a mean of 3.25 inch, together their mean won't be 3.25 inch, but 6.5 inch.

    What you need to know is that the sum of 2 independent normal distributions is again a normal distribution.
    Its mean is the sum of the means, and its variance is the sum of the variances.
     
    Last edited: Oct 1, 2011
  8. Oct 1, 2011 #7
    So I should compute: [tex] P(0 \le x \le 6.55) = P( \frac{0-6.5}{2 \sqrt{.008}} \le Z \le \frac{6.55-6.5}{2 \sqrt{.008}}) = P(-36.336 \le Z \le 0.2795) [/tex] ?

    Does that mean I should calculate (z=0.2795) = 61% ? (approximately, I'll use my real calculator and more detailed z table later)
     
    Last edited: Oct 1, 2011
  9. Oct 1, 2011 #8

    vela

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    Oops! I guess I should have used y.
     
  10. Oct 1, 2011 #9

    I like Serena

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    Yeah, I mixed up the use of z, especially since Arcana did not define Z herself. :wink:


    I was just going to answer that you had it right...
    But then you changed it, and now it's not right anymore. :frown:
     
  11. Oct 1, 2011 #10
    Crap. But if [itex] \sigma ^2 = .008 [/itex] then [itex] \sigma = 0.08944 [/itex] so I thought to add [itex] \sigma + \sigma = 2 \sqrt{ \sigma ^2} [/itex], not [itex] \sqrt{ 2 \sigma^2} [/itex]
     
  12. Oct 1, 2011 #11

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    Yeah, well, you need to add the variances (which you did) and not the standard deviations.
     
  13. Oct 1, 2011 #12
  14. Oct 1, 2011 #13

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    Hmm, you may want to take a look here:
    http://en.wikipedia.org/wiki/Sum_of_normally_distributed_random_variables

    It explains how you add 2 independent normal distributions and it also gives 3 proofs why this is so.


    More intuitively you can think of it as follows.

    In the worst case scenario both cylinders are longer than they should be, or they are both shorter than they should be.
    In the worst case scenarios the standard deviations would add up (this is when the lengths of the cylinders are "correlated" and not independent).

    However, in reality there is a good chance that when one cylinder is longer the other is shorter, meaning the new standard deviation will be less than the sum of the 2 standard deviations.

    It turns out that in reality the variances add up instead of the standard deviations.
     
  15. Oct 1, 2011 #14
    So it would be close to 65% like I said before I changed it?
     
  16. Oct 1, 2011 #15

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    Yep! :smile:
    It would be 65.3%.
     
  17. Oct 1, 2011 #16
    Thank you very much! :)
     
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