SOLUTION: I am having problems figuring out where to even start with the following, even just a hit would be appreciated. According to a survey, College students make an average of 18 ca

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Question 909169: I am having problems figuring out where to even start with the following, even just a hit would be appreciated.
According to a survey, College students make an average of 18 calls per day on their cell phone. Moreover, 85% of the students surveyed indicated that their parents pay their cell phone expenses.
a) If you select a student at random, what is the probability that he or she makes more than 10 calls in a day? Less than 15 calls a day? At least 20 calls a day?
b) If you select a random sample of 15 students, what distribution can you use to model the proportion of students who have parents who pay their cell phone expenses?
c) Using the distribution selected in (b), what is the probability that all 15 have parents who pay their cell phone expenses? At least 10?

Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
a) Poisson Distribution (average = 10 per day):
P( x >10) = 1 - poissoncdf(10,10)
P( x < 15) = poissoncdf(10,14)
P(x ≥ 20) = 1 - poisson(10,19)
b) Binomial Distribution (parents pay 0r don't pay)
p(pay) = .85, n = 15
c)P(x = 15) = binompdf(15, .80, 15) = .85^15
P(x ≥ 10) = 1 - binomcdf(15, .85, 9)
Using P+%28x%29=+highlight_green%28nCx%29%28p%5Ex%29%28q%29%5E%28n-x%29+ = 15C15(.85)^15(.15)^0
p and q are the probabilities of success and failure respectively.
In this case p= .85, q = .15 , n = 15
nCx+=+%28n%21%29%2Fx%21%28n+-+x%29%21%29