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Need help with statistics

  1. Aug 13, 2010 #1
    1. The problem statement, all variables and given/known data
    the weight of items produced by a production line is normally distributedwith 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!)) * [tex]\pi[/tex]^k * (1-[tex]\pi[/tex])^(n-k)
    mean = n[tex]\pi[/tex]
    variance = n[tex]\pi[/tex](1-[tex]\pi[/tex])

    if x is N(µ,[tex]\sigma[/tex]), then z=(x-µ)/([tex]\sigma[/tex]) is N(0,1)

    3. The attempt at a solution
    a. x~n (µ=12, [tex]\sigma[/tex]=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 (µ=7, [tex]\sigma[/tex]=.95)
    p(x=3)
    p(x=K)=[tex](3!)/(3!)(7-3)![/tex](.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
    (µ=12, [tex]\sigma[/tex]=2)
    P(x<10) or P(x>14)
    Hope you people can help me
     
  2. jcsd
  3. Aug 14, 2010 #2

    danago

    User Avatar
    Gold Member

    The events X>14 and X<10 are mutually exclusive (cannot occur at the same time) so that their probabilities are additive:

    P(X>14 or X<10) = P(X>14) + P(X<10)
     
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