An article reported that what airline passengers like to do

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 90% did so. Suppose that for a particular route the actual percentage is exactly 90%, 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 = 0.9. (Round your answers to four decimal places.)

Can you show all the steps please?

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

Interpret this probability.

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.    

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

This is the probability that exactly 6 out of 10 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) =

Solution

Binomial Distribution

PMF of B.D is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
Where   
k = number of successes in trials
n = is the number of independent trials
p = probability of success on each trial

B~(n,p), We have B~(9,0.90)
a)

P( X = 6 ) = ( 9 6 ) * ( 0.9^6) * ( 1 - 0.9 )^3
= 0.0446
This is the probability that exactly 6 out of 9 selected passengers rested or slept.

b)
P( X = 9 ) = ( 9 9 ) * ( 0.9^9) * ( 1 - 0.9 )^0
= 0.3874

c)

P( X < 6) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= ( 9 5 ) * 0.9^5 * ( 1- 0.9 ) ^4 + ( 9 4 ) * 0.9^4 * ( 1- 0.9 ) ^5 + ( 9 3 ) * 0.9^3 * ( 1- 0.9 ) ^6 + ( 9 2 ) * 0.9^2 * ( 1- 0.9 ) ^7 + ( 9 1 ) * 0.9^1 * ( 1- 0.9 ) ^8 + ( 9 0 ) * 0.9^0 * ( 1- 0.9 ) ^9
= 0.0083
P( X > = 6 ) = 1 - P( X < 6) = 0.9917