An article reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost 80% did so. Suppose that for a particular route the actual percentage is exactly 80%, and consider randomly selecting nine passengers. Then x, the number among the selected nine who rested or slept, is a binomial random variable with n = 9 and p = 0.8. (Round your answers to four decimal places.)

(a) Calculate p(6).
p(6) =

Interpret this probability.

This is the probability that exactly 6 out of 10 selected passengers rested or slept.This is the probability that at least 6 out of 9 selected passengers rested or slept. This is the probability that at least 6 out of 10 selected passengers rested or slept.This is the probability that exactly 6 out of 9 selected passengers rested or slept.

(b) Calculate p(9), the probability that all nine selected passengers rested or slept.
p(9) =

(c) Determine P(x ≥ 6).
P(x ≥ 6) =

The sample taken is n=9 passengers and the probability of success, that is finding a passenger that either rested or sept during the flight, is p=0.80.

I'll use the binomial tables to calculate each probability, these tables give the values of accumulated probability: P(X≤x)

a. P(6)= P(X=6)

To reach the value of selecting exactly 6 passengers you have to look for the probability accumulated until 6 and subtract the probability accumulated until the previous integer:

P(X=6)= P(X≤6)-P(X≤5)= 0.2618-0.0856= 0.1762

b. P(9)= P(X=9)

To know the probability of selecting exactly 9 passengers that either rested or slept you have to do the following:

P(X≤9) - P(X≤8)= 1 - 0.8657= 0.1343

c. P(X≥6)

To know what percentage of the probability distribution is above six, you have to subtract from the total probability -1- the cumulated probability until 6 but without including it:

P(X≥6)= 1 - P(X<6)= 1 - P(X≤5)= 1 - 0.0856= 0.9144

I hope it helps!