document.write( "Question 909169: I am having problems figuring out where to even start with the following, even just a hit would be appreciated.\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "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?\r
\n" ); document.write( "\n" ); document.write( "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?\r
\n" ); document.write( "\n" ); document.write( "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?
\n" ); document.write( "
\n" ); document.write( "

Algebra.Com's Answer #551703 by ewatrrr(24785)\"\" \"About 
You can put this solution on YOUR website!
a) Poisson Distribution (average = 10 per day):
\n" ); document.write( "P( x >10) = 1 - poissoncdf(10,10)
\n" ); document.write( "P( x < 15) = poissoncdf(10,14)
\n" ); document.write( "P(x ≥ 20) = 1 - poisson(10,19)
\n" ); document.write( "b) Binomial Distribution (parents pay 0r don't pay)
\n" ); document.write( "p(pay) = .85, n = 15
\n" ); document.write( "c)P(x = 15) = binompdf(15, .80, 15) = .85^15
\n" ); document.write( "P(x ≥ 10) = 1 - binomcdf(15, .85, 9)
\n" ); document.write( "Using \"P+%28x%29=+highlight_green%28nCx%29%28p%5Ex%29%28q%29%5E%28n-x%29+\" = 15C15(.85)^15(.15)^0
\n" ); document.write( "p and q are the probabilities of success and failure respectively.
\n" ); document.write( "In this case p= .85, q = .15 , n = 15
\n" ); document.write( "\"nCx+=+%28n%21%29%2Fx%21%28n+-+x%29%21%29\"
\n" ); document.write( "
\n" ); document.write( "
\n" );