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Normal Distribution

  1. Jul 4, 2011 #1
    1. The problem statement, all variables and given/known data

    A company specifies that the actual contents of packages shall not be less, on average, than the Nominal Quantity (Qn) and that not more than 2.5% of packages may be Non-Standard (NS), i.e. deviate from Qn by more than "total negative error" (TNE). Thus, if Qn=230g and TNE=9g a package is NS if it weighs less than 221g.
    Assume a normal distribution and containers are filled independently.

    (a) Let Qn=230g and TNE=9g and suppose the company sets the average fill to be u=230g.

    (i) What must the standard deviation of the filling operation be so that only
    2.5% of containers are NS?

    (ii) Suppose the actual standard deviation is 6g, what minimum value should u be so that
    the NS regulation is met? Round to whole units.


    (b) Suppose u is the value found in (a)(ii) and tha five containers are selected at random
    and weighed.

    (i) What is the probability that two or more of them are NS?

    (ii) What is the probability that the combined weight of the five congtainers is less than
    1200g?


    (c) What is the probability that an inspector will need to weigh exactly 6 containers
    from the production line to discover a NS container. That is, the first five are standard
    and the sixth is NS.


    2. Relevant equations



    3. The attempt at a solution

    I don't have my actual workings with me, but here are my methods.


    (a) (i) 2.5% is NS, so I found the z-value, Z, which gives 0.025.
    Z = (221-230)/SD

    From this I found the standard deviation.

    (ii) Z = (221 - Umin)/6
    Solved for Umin


    (b) (i) P(2 or more are NS) = 1 - P(1 or 0 is not NS)
    How do I calculate this probabilty?

    (ii) I have no idea here


    (c) Do I have to do a binomial distribution?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Jul 4, 2011 #2

    I like Serena

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    There you are again! :smile:

    Your method is good, your z-value is wrong.
    (A typo I presume?)


    What is the chance that exactly 0 containers are NS?
    What is the change that only the 1st container is NS?
    What is the change that only the 2nd container is NS?

    Can these chances be added up?


    Alternatively, what do you know about the binomial distribution?
    And in particular the cumulative binomial distribution?
    Can you think of parameters that would be applicable?


    If you have 2 normal distributions and you add them up.
    What is the resulting mean?
    And what is the resulting variance?

    What if you add 5 normal distributions?


    No, not in this case.

    This would be:
    P(first ok AND second ok AND third ok AND fourth ok AND fifth ok AND sixth is NS)

    Do you know how to calculate this?
     
  4. Jul 5, 2011 #3
    Sorry, that was a typo. Should have been z = -1.96
    Standard deviation = 4.5918

    Umin = 233g

    P(chosen bottle is NS) = 0.025

    So the desired probability is
    1 - 5C0.(0.025)^0.(0.975)^5 - 5C1.(0.025)^1.(0.975)^4 - 5C2.(0.025)^2.(0.975)^3
    = 0.00015045

    I'm not really sure here..

    Is it (0.975)^4.(0.025) ?
     
  5. Jul 5, 2011 #4

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    Good! :smile:


    You're in the neighbourhood.
    However, you switched your chances around, since the chance on not NS is 0.975.
    And you should leave out the last term, since that is for 2 bottles! :wink:

    Btw, you're effectively looking at a binominal distribution here, with parameters n=5 and p=0.975.
    So you could also calculate it with the cumulative binomial distribution.
    That is: P(2 or more are NS) = 1 - P(1 or 0 is not NS) = 1 - binomcdf(n=5, p=0.975, k=1)


    If you add up 2 independent normal distributions, you'll get a new normal distribution, with:
    1. a mean that is the sum of the two means
    2. a variance, that is the sum of the two variances
    (See wikipedia: http://en.wikipedia.org/wiki/Sum_of...andom_variables#Independent_random_variables").

    Can you continue this?


    Almost, your reasoning is sound, but I think you miscounted.
    The first power should be 5, since their are 5 bottles ok.
     
    Last edited by a moderator: Apr 26, 2017
  6. Jul 5, 2011 #5
    When I did it I was putting getting a NS as a success and a standard as a failure. I understand your way, but not why mine is wrong.

    For 5 items the mean would be the sum of the 5 means.
    The variance would be the sum of the 5 variances.

    Each item has U=233, SD=6, Variance=36
    So for the sum, U=1165, SD=13.4164

    Normalize 1200, and get the corresponding probability.
    Is this right?

    [/QUOTE]

    Sorry, that was a type. (0.975)^5.(0.025)
    Do I have to do anything else to this? Isn't this the same as first 4 are ok, 5th is NS, 6th is ok?
     
    Last edited by a moderator: Apr 26, 2017
  7. Jul 5, 2011 #6

    I like Serena

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    You calculated
    [itex]1 - \binom 5 0 (0.025)^0 (0.975)^5 - \binom 5 1 (0.025)^1 (0.975)^4 - \binom 5 2 (0.025)^2 (0.975)^3[/itex]​

    which is:
    1 - P(0 bottles NS) - P(1 bottle NS) - P(2 bottles NS) = P(3, 4 or 5 bottles NS)​

    but that was not the question!

    Yep! :smile:


    That is correct!
    And yes it is the same.
     
  8. Jul 5, 2011 #7
    I had a typo, I should have written
    [itex]1 - \binom 5 0 (0.025)^0 (0.975)^5 - \binom 5 1 (0.025)^1 (0.975)^4[/itex]​

    This gives
    1 - P(0 bottles NS) - P(1 bottle NS) = P(2, 3, 4 or 5 bottles NS)​

    The question asked "what is the probability at least two of them are NS, so isn't this correct?
     
  9. Jul 5, 2011 #8

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    Yes it is.
    Just checking if you were still sharp! :smile:

    (It seems I switched the chances around myself. :shy:)
     
  10. Jul 5, 2011 #9
    Okay! :smile:

    Thank you very much for all your help! :smile:
     
  11. Jul 5, 2011 #10

    I like Serena

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    Appreciate the thanks!
    And see you next time. :cool:
     
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