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Math Help - Binomial Distribution (LEVEL 2 CALCULUS)

  1. #1
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    Smile (SOLVED) Binomial Distribution (LEVEL 2 CALCULUS)

    Question:

    A hundred samples of 5 resistors were taken from a large batch. 59 samples had no defective resistors, 33 had 1, 7 had 2, 1 had 3, and no samples had 4 or 5 defective resistors. Show that the distribution is approximately binomial and estimate the overall percentage of defective components.

    I need some help in determining the variables p,q,n to solve this. Please correct me if I'm wrong:

    q = possibility of no defect in a sample = 1/5
    p = possibility of having defects in a sample = 4/5
    n (times) = 100

    ANSWER of overall percentage = about 10%
    Last edited by Salcybercat; April 25th 2011 at 01:17 AM. Reason: Question is solved
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  2. #2
    Senior Member Sambit's Avatar
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    n (times) = 100
    This is wrong. Here n=number of defects in each sample.
    Show that the distribution is approximately binomial...
    I think this means you have to FIT the given distribution with a binomial distribution having suitable parameters. To do this, first compute the average number of defectives from the given samples (the formula is sum of (value*frequency) / sum of frequencies which will serve as the "possibility of no defect" that is p. So now you have a binomial distribution of parameters (n=100,p). From this, calculate the probabilities of X=0,1,2,3,4,5
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  3. #3
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    So now you have a binomial distribution of parameters (n=100,p)
    I'm a bit unclear regarding this. Earlier you said it's wrong? Below is some of the workings with your help. There was a typo in my earliest post: p = probability of having defect and q = probability of having no defect.
    Do take a look and comment on the binomial equation I came up with, please.

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  4. #4
    Senior Member Sambit's Avatar
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    I apologize for two reasons:
    I'm a bit unclear regarding this. Earlier you said it's wrong?
    I meant to say you have a binomial distribution of parameters (n=5,p)
    And the second one is : the formula for getting p is { sum of (value*frequency) / sum of frequencies } / n so it will be 0.1 (divide your result by n that is by 5).

    Apart from this, your method is correct. Sorry or the confusion; I am getting mad at my browser which is making a lot of trouble now-a-days.
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  5. #5
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    I got the answer! Thank you very much for your help
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