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Homework Help: Stat #3

  1. Feb 13, 2006 #1
    Here is what I know,

    It is good if [tex] X=x=R \leq 2 [/tex]
    n = 100
    p=1% is rated Good
    p=5% is rated bad

    Im guessing I have to use a Binomial distribution here??

    [tex] p(x) = P(X=x) = \frac{ n!}{x!(n-x)!} p^x(1-p)^{n-x} [/tex]

    So,

    [tex] X\sim Bin(100,0.01) [/tex]

    The fact that we want 2 or fewer welds means

    [tex] P(X \leq 2) = P(X=0-or-X=1-or-X=2) [/tex]

    PART A)

    The probability that a good deisgn will be rejected will be equal to:

    [tex] 1 -p(x) = 1- P(X=x) = 1 -\frac{ n!}{x!(n-x)!} p^x(1-p)^{n-x} [/tex]

    [tex] 1 -\frac{ 100!}{x!(100-x)!} (.01)^x(1-.01)^{100-x} [/tex]

    for x=0,1,2:

    [tex] P(X=x \leq 2) = P(x=0)+P(x=1)+P(x=2) [/tex]

    Because of the EXCESSIVE n value, its not going to work on my Ti-83 calculator and I don't feel like busting out matlab. So I am going to use the cumulative table provided by the prof online that I just found by mistake (nice of him to mention needing to use it :grumpy:)

    [tex] p(x \leq 2)=0.92063 [/tex]

    So the probability that a good design gets rejected is

    [tex] 1 - p(x \leq 2) = 0.07937 [/tex]

    which is equivalent to, 7.937% good parts accidentially rejected.

    PART B)

    The same thing as Part a, however the table now refers to values for

    [tex] X\sim Bin(100,0.05) [/tex]

    [tex] p(x) = P(X=x) = \frac{ n!}{x!(n-x)!} p^x(1-p)^{n-x} [/tex]

    [tex] \frac{ 100!}{x!(100-x)!} (.05)^x(1-.05)^{100-x} [/tex]

    for x=0,1,2:

    [tex] P(X=x \leq 2) = P(x=0)+P(x=1)+P(x=2) [/tex]

    [tex] p(x \leq 2)=0.011826 [/tex]

    so

    1.1826% are accepted from the bad batch.
     
    Last edited: Feb 13, 2006
  2. jcsd
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