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.) Required:
a. Calculate p(4).
b. Calculate p(6), the probability that all six selected passengers rested or slept.
c. Determine P(x ≥ 4).
d. Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below.

1. P(z < 2.36) =
2. P(z lteq. gif 2.36) =
3. P(z < -1.23) =
4. P(1.15 < z < 3.35) =
5. P(-0.77 lteq. gifz lteq. gif -0.54) =
6. P(z > 3) =
7. P(z gteq. gif -3.48) =
8. P(z < 4.97) =