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  1. #1
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    need help with statistics




    1. The problem statement, all variables and given/known data
    the weight of items produced by a production line is normally distributed with a mean of 12 ounces and a standard deviation of 2 ounces.
    a. what is the probability that a randomly selected item will weight between 8 and 16 ounces? (DONE)
    b. what is the probability that a randomly selected item will weight over 20 ounces? (DONE)
    c. Suppose that quality control requires the weight of items to be within 8 and 16 ounces. You select 7 items at random (each item is independent). What is the probability that 3 of the items will fulfill quality control requirements. (DONE BUT HAVE DOUBTS)
    d. find the probability that a randomly selected item has a weight that is greater than 14 or smaller than 10. (STUCK IN THIS ONE)

    2. Related formulas
    if x is BIN
    p(x=k) = (n!)/((n-k)!(k!)) *(pi)^k * (1-pi)^(n-k)
    mean = n(pi)
    variance = n(pi)(1-(pi))

    if x is N(µ,σ), then z=(x-&#181/(σ) is N(0,1)

    3. The attempt at a solution
    a. x~n (µ=12, σ=2)
    p(8<x<16)
    p(x<16) - p(x<8)
    z=(16-12)/2 z=(8-12)/2
    z=2 z= -2
    p(z<2)-p(z<-2)
    =.9772-.0228
    =.9544

    b.P(x>20)
    z=(20-12)/2
    z=4
    p(z>4)= 1

    c.x~binomial (&#181;=7, =.95)
    p(x=3)
    p(x=K)= (3!)/((3!)(7-3)!)*(.95)^3 *(1-.95)^4
    .
    .
    .
    x=0.000187551

    d. I have no idea how to deal with this one
    I think I have to use the mean and standard deviation of the problem
    (&#181;=12, =2)
    P(x<10) or P(x>14)
    Hope you people can help me
    Last edited by 619snake; 08-13-2010 at 02:23 PM.

  2. #2
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    Quote Originally Posted by 619snake View Post
    [B]
    c.x~binomial (=7, =.95)
    p(x=3)
    p(x=K)=(.95)^3 (1-.95)^4
    .
    .
    .
    x=0.000187551
    You're final solution looks correct to me but you forgot the 7!/(3!4!) constant out front when you were deriving the answer.
    d. I have no idea how to deal with this one
    I think I have to use the mean and standard deviation of the problem
    (=12, =2)
    P(x<10) or P(x>14)
    Hope you people can help me
    An easier way to do problems like these is to find the probability of getting the complement of what you want ( 10<x<14 ) and then we know that if A and A' are complements then P(A) = 1 - P(A')

  3. #3
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    thanks a lot for your help, I think I've figured it out!

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