Question

The Los Angeles Times (December 13,1992 ) 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 \(\pi=.8\). a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6)\), the probability that all six selected passengers rested or slept. c. Determine \(P(x \geq 4)\).

Short answer

a. The probability \(P(4)\) gives the likelihood that exactly 4 out of 6 passengers will rest or sleep on a specific route. b. The probability \(P(6)\) gives the likelihood that all 6 passengers will rest or sleep on a specific route. c. The probability \(P(x \geq 4)\) gives the likelihood that 4 or more passengers will rest or sleep on a specific route. Detailed solutions can be found above.

Step-by-step solution

Identify Known Variables

Firstly, the parameters of the problem need to be identified: number of attempts \(n=6\) and probability of success \(\pi=0.8\).

Calculate Probability for 4 Passengers

Apply the binomial probability formula, where \(x\) is the number of successes (e.g., passengers who rest or sleep), which in this case is 4: \[P(4) = C(6,4) \cdot (0.8)^4 \cdot (0.2)^2\] where \(C(6,4)\) represents the combinations of 6 items taken 4 at a time.

Calculate Probability for 6 Passengers

Use the binomial probability formula again, but this time for \(x=6\): \[P(6) = C(6,6) \cdot (0.8)^6 \cdot (0.2)^0\]

Calculate Cumulative Probability

Now the probability that 4 or more passengers will rest or sleep has to be calculated. This is equivalent to the sum of the probabilities of 4, 5, and 6 successes: \[P(x \geq 4) = P(4) + P(5) + P(6)\] where \(P(5) = C(6,5) \cdot (0.8)^5 \cdot (0.2)^1\). By calculating these values an approximate solution can be found.