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Rate 5 stars quick, but please provide detailed working so i can understand

Question 1 Airlines usually over-book the seats on an aircraft by a certain margin because they know from experience that some people change or do not show for their scheduled flight. Data collected for a particular Melbourne-Darwin flight showed that, on average, 240 people (with a standard deviation of 30) did arrive for their scheduled flight. This data was distributed normally. The aircraft has seats for 310 passengers. a) What are the z-scores for 150, 210 and 250 arrivals? b) What is the probability that a randomly selected flight has enough seats available for all the people who turn up? c) This particular flight goes 7 days a week, 52 weeks a year. During one year of operation, how many times would there be more passengers than available seats?

Explanation / Answer

a) z = (x - Mean)/S.D.

For x=150,

z = (150 - 240)/30 = -3

For x=210,

z = (210 - 240)/30 = -1

For x=250,

z = (250 - 240)/30 = 0.33

b) Total no of seats = 310

For x = 310,

z = (310 - 240)/30 = 2.33

From normal distribution table, probability = 0.99

c) Probability that passengers are more than no of seats = 1 - 0.99 = 0.01

No of times this happens in a year = 0.01 * Number of flights

Number of flights =7 * 52 = 364

Number of times = 364 * 0.01 = 3.64 ~ 4

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