11. An airline overbooks a flight by selling 100 tickets for a flight that has o
ID: 3336703 • Letter: 1
Question
11. An airline overbooks a flight by selling 100 tickets for a flight that has only 95 seats available. From past experience with this type of flight, the airline estimates that the probability that a person with a ticket actually shows up f flight. If this is so, they will have to offer some incentive such as free tickets to a future flight in order to induce one or more of the ticket holders to agree to wait for a later flight. Let X represent the number of ticket holders that actually arive for the flight. Using the binomial distribution, the probability that more than 95 for the flight is .90. The airline is concerned that more than 95 ticket holders may arrive for the ticket holders will arrive for the flight can be modeled as b) n-100, -95, X > 90 c) n-100, 90, x>95 d) n-95, .90, X > 100 e) n-100, -90, X> 100 12) All of the following conditions must be met in order to apply the binomial distribution except a) the probability of success must be 50 for every observation b) separate observations must be independent c) the scenario under study can be thought of as consisting of two possible outcomes d) the number of opportunities for success (observations) must be known Part 2 Problems 1. (9 pts) Wage data collected by a labor economist indicate that the hourly wage of wel is approximately normally distributed with a mean of S34 and a standard deviation of What is the probability that a welder earns between $30 and $40 per hour? (8 pts) Completion times for a civil service exam follow a normal distribution w an of 95 minutes and a standard deviation of 20 minutes. What proportion of those ta exam will need more than 2 hours (120 minutes) to complete the exam?Explanation / Answer
(11) Given X : a random variable denoting the number of ticket holders that actually arrive for the flight.
To find, the probability that more than 95 ticket holders will arrive for the flight, that is, P(X > 95).
If X follows a Binomial distribution, then n = number of trials = number of tickets sold by the airline = 100.
Also, pi = probability of success = probability that a person with a ticket will actually turn up for the flight = 0.90
Hence, correct answer is Option C.
(12) To apply the binomial distribution, the separate observations must be independent, there are only two possible outcomes - failure or success and the number of oppurtunities for success (in terms of observations) must be known from beforehand.
Hence, correct answer is Option A, since the probability of success for every observation is not only restricted to 0.50, but also it can be of any value between 0 and 1. For example, in the above problem, the probability of success was 0.90.
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