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 six passengers. Then x, the number among the selected six who rested or slept, is a binomial random variable with n = 6 and p = 0.8. (Round your answers to four decimal places.)

(a) Calculate p(4).
p(4) =

Interpret this probability.

This is the probability that exactly 4 out of 6 selected passengers rested or slept. This is the probability that exactly 4 out of 10 selected passengers rested or slept. This is the probability that at least 4 out of 6 selected passengers rested or slept. This is the probability that at least 4 out of 10 selected passengers rested or slept.
(b) Calculate p(6), the probability that all six selected passengers rested or slept.
p(6) =
(c) Determine P(x ≥ 4).P(x ≥ 4) =

Let z denote a random variable that has a standard normal distribution. Determine each of the probabilitiesbelow. (Round all answers to four decimal places.)

(a) P(z < 2.36) =
(b) P(z lteq. gif 2.36) =
(c) P(z < -1.23) =
(d) P(1.15 < z < 3.35) =
(e) P(-0.77 lteq. gifz lteq. gif -0.54) =
(f) P(z > 3) =
(g) P(z gteq. gif -3.48) =
(h) P(z < 4.97) =