An article reported that what airline passengers like to do most on long flights

Question

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 six passengers. Then x, the number among the selected six who rested or slept, is a binomial random variable with n = 6 and p = 0.9. (Round your answers to four decimal places.)

(a) Calculate p(4). p(4) = Interpret this probability. This is the probability that at least 4 out of 10 selected passengers rested or slept. 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.

(b) Calculate p(6), the probability that all six selected passengers rested or slept. p(6) =

(c) Determine P(x 4). P(x 4) =

Explanation / Answer

This is the probability that exactly 4 out of 6 selected passengers rested or slept.

P(x=4 ) = 6C4 * 0.90^4 * 0.10^(6-4) = 0.098415

P( x = 6) = 6C6 * 0.90^6 * 0.10 ^[6-6] = 0.531441

P( x >= 4 ) = P( x=4) + P(x=5) + P(X=6)

P(x=5) = 6C5 * 0.90^5 * 0.10 ^[6-5] = 0.354294

P( x >= 4 ) =0.098415 + 0.354294 + 0.531441 = 0.98415

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