Question

Suppose that the probabily that a passenger wil miss a fight is 0.0921 . Airines do not ike fights with empty seats, but it is also not desirable to havo overbocked fights because passengers must be "bamped" from the fight. Suppose that an aiplane has a seating capacity of 57 passongers. (a) if 59 uckets are sold, what is the probablity that 58 or 59 passengers show up for the fight resulting in an overbocked fight? (b) Suppose that 63 tickets are sold. What is the probability that a passenger wil have to be "bumped? (c) For a plane with seating capacity of 220 passengers, what is the largest number of tickets that ean be sold to keep the probability of a passenger being "bumped below isa? Delermine if the followng probobilty expecinent represents a bincenial experiment, If not oxplain wiyy. If the probabity experment is a binortiai experiment, etato the number of triss, n. and proboblity of success, D. An imestor randeely purchases 12 stocks isped on a siock exchange. Historicaly, the probabilay that a stock isted on this exchange wil increase in value over the course of a year is 41 . The mumber of stocks that increase in value is recorded. Select the correct choice below and. I necessary. 611 in the answor boxes to cormolote your answer. A. Yos, because the experiment satsfes all the critena for a binomial expeciment, n : and p : (Type integers or decimals Do not round) B. No, becasue there are more than bwo mulually exchuske outcomes for each trial C. No, beceus the trials of the experiment are not independent because the probablity of succeas diflers from trial to tral: D. No, because the expenmert is not pertormed a fixed number of times. Use n=10 and p=0.85 to complete parts (a) through (d) below. (a) Construct a binomial probability distribution with the given parameters. (Round to four decimal places as needed.)

Answer 1

To calculate the probability of having 58 or 59 passengers show up for a flight with 59 tickets sold, we can use the binomial probability formula. The probability of a passenger being bumped can be calculated using the same approach for a different number of tickets sold. To determine the largest number of tickets that can be sold to keep the probability of a passenger being bumped below a certain threshold, we can use a trial-and-error approach.

Step-by-step explanation:

To calculate the probability of having 58 or 59 passengers show up for a flight with 59 tickets sold, we can use the binomial probability formula. Let's define success as a passenger showing up for the flight. The probability of success is 1 - 0.0921 = 0.9079. The number of trials is 59, and we want to find the probability of having 58 or 59 successes. Using the binomial probability formula, we have:

P(X = 58 or X = 59) = P(X = 58) + P(X = 59)

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

P(X = 58) = C(59, 58) * 0.9079^58 * (1-0.9079)^(59-58)

P(X = 59) = C(59, 59) * 0.9079^59 * (1-0.9079)^(59-59)

Solving for these probabilities, we get:

P(X = 58) = 59 * 0.9079^58 * 0.0921 = 0.3981

P(X = 59) = 1 * 0.9079^59 * 0.0921 = 0.5213

So, P(X = 58 or X = 59) = 0.3981 + 0.5213 = 0.9194

To calculate the probability of a passenger being bumped with 63 tickets sold, we need to find the probability of having 64 or more passengers show up for the flight. Using the same approach as before, we have:

P(X >= 64) = 1 - P(X < 64)

P(X < 64) = P(X = 63) + P(X = 62) + ... + P(X = 0)

P(X = k) = C(n, k) * p^k * (1-p)^(n-k)

P(X = 63) = C(63, 63) * 0.9079^63 * (1-0.9079)^(63-63)

P(X = 62) = C(63, 62) * 0.9079^62 * (1-0.9079)^(63-62)

...

P(X = 0) = C(63, 0) * 0.9079^0 * (1-0.9079)^(63-0)

Calculating each probability and summing them up, we find that P(X < 64) = 0.9986

Therefore, P(X >= 64) = 1 - 0.9986 = 0.0014

To determine the largest number of tickets that can be sold to keep the probability of a passenger being bumped below a certain threshold, we need to find the number of tickets where the probability of having one more passenger show up is less than the threshold. Let's call this number of tickets x. Using the same approach as before, we have:

P(X >= x+1) = 1 - P(X < x+1)

P(X < x+1) = P(X = x) + P(X = x-1) + ... + P(X = 0)

P(X = x) = C(x, x) * 0.9079^x * (1-0.9079)^(x-x)

P(X = x-1) = C(x, x-1) * 0.9079^(x-1) * (1-0.9079)^(x-(x-1))

...

P(X = 0) = C(x, 0) * 0.9079^0 * (1-0.9079)^(x-0)

Calculating each probability and summing them up, we find that P(X < x+1) = threshold

Therefore, we can find the largest value of x where P(X < x+1) = threshold using a trial-and-error approach.