given 58% pay late fees, find prob. that, in random 20 accts

[csing1019]

Hello. I am dyer need of some assistance. I have two problems I need help with and I can not figure them out. I have tried hard and long and I have called several friends and family but we all have been out of school too long and can not figure out how to come up with the answer. The question is:

58% of all credit card customers pay late fees. If a random sample of 20 credit card holders is selected, what is the probablility that

a.0 had to pay a late fee.
b.no more than 5 had to pay a late fee
c.more than 10 had to pay late fees
d.what assumptions were made while answering a-c

My first instict was to simply subtract 100-58 but from my research I need to do some dividing. I'm not sure what I should actually be dividing or doing.

Thank you in advance for your assistance.

Candice

 

[arthur ohlsten]

Re: Question regarding Statistic Probablility

I do not know what you are studying. I will do it one way that does not assume a normal distribution but rather a binomial distribution

Probability customer pays a late fee P[f]=.58
probability customer does not pay a late fee Q[f]=.42; this of course is 1 - P[f]

the distribution of a random sample of 20 customers is:
[P[f]+Q[f]]^20= P^20 + 20 P[f]^19Q[f] +.....Q[f]^20

general term is 20CmP[f]^mQ[f]^(20-m)
where the coefficients from combinational notation is
nCm=n / [m![n-m]! ] the combination of n things taken m at a time
or the coefficients are from pascals triangle

a) 0 late fee is P[f]^0Q[f]^20
[.42]^20 answer

b) 20C5 P^5Q^15+ 20C4 p^4Q^16+20C3 p^3q^17+ 20C2 p^2q^18+ 20pq^19+1 q^20


c) is sum of p^20 term down to p^10 term

d0 is I assumed a binomial distribution

Hope this helps
please check for errors
Arthur