SOLUTION: Eighty percent of (80%)of business travelers carry either a cell phone or a laptop. Assume a binomial probability distribution for the number x of business travelers carrying a cel

 

Hi,

Note: The probability of x successes in n trials is: 

P = nCx*  where p and q are the probabilities of success and failure respectively. 

In this case p(carry either a cell phone or a laptop) = .8 & q = .2  and  n = 15

A)what are the possible values of the number of business travelers carrying a cell phone or a laptop in the sample? 

 0 ≤  x  ≤   15

B)What is the expected number of business travelers carrying a phone or a laptop in such a sample? 

C) Is this probability distribution skewed to right or left?

Note:  distributions with fewer observations on the left (toward lower values) are said to be skewed left.

D)Compute the standard deviation of the number of business travelers carrying a phone or a laptop in such a sample. 

E)What is the probability that exactly 8 business travelers in a sample carry a cell phone or a laptop?

TI Calculator: binompdf(n, p, x-value) = binompdf(15, .8,8) 

F)What is the probability that 7 of the 15 business travelers have neither a cell phone nor a laptop? binompdf(15, .2,7) 

G)What is the probability that at least 10(ten) business travelers in the sample have a cell phone or a laptop? 

1 – binomcdf(n, p, largest x-value)= 1 – binomcdf(15, .8, 9)