# Statistics, Probability Distribution

1. Oct 18, 2007

### kuahji

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. Oct 18, 2007

### EnumaElish

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
3. Oct 18, 2007

### kuahji

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...

4. Oct 19, 2007

### kuahji

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 :).