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Statistics, Probability Distribution

  1. Oct 18, 2007 #1
    I have the data set
    # of freights - probability
    1 - .075
    1.5 - .025
    2 - .425
    2.5 - .150
    3 - .125
    3.5 - .100
    4 - .050
    5 - .025
    6 - .025

    The question is, what is the probability that at least 5 of 6 families purchased more than four freights.

    I started with making a graph of the problem & found it was skewed right. With a mean 2.537 & standard deviation 1.002. So I've ruled out using normal distribution because its skewed right. The probability of purchasing more than four freights is .025+.025=.050. But is there a way to calculated the probability that 5 of 6 families purchased more than four?
     
  2. jcsd
  3. Oct 18, 2007 #2

    EnumaElish

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    How many ways can you come up with in which 5 out of 6 families (A, B, C, D, E, and F) purchase at least 4? For example, each family makes 1 purchase. Or 5 families make 1 each. Or one family makes 4, 5, or 6... etc.
     
    Last edited: Oct 18, 2007
  4. Oct 18, 2007 #3
    There should be 216 different ways the 6 families buy between 4 & 6 freights. Take 6^3 to get that result. Then 5^3 should give 125, for 5 families. Or at least I think...
     
  5. Oct 19, 2007 #4
    Just ended up trying to solve it like I would a binomial experiment. Where I added up the probability of 4 or greater which equaled .1. Then plugged it into the calculator as bnpfd(6,.1,5) to get an answer of 5.4^-5 & dubbed check with the long hand equations. Couldn't think of any other way to solve the problem. But I figured it was either they did or did not purchase 4 or more freights, each person does not effect the other, & probability of success/failure is the same. So hopefully that works, guess I'll see soon enough :).
     
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