Airlines sometimes overbook flights. Suppose that for a plane with 40 seats, 43

Question

Airlines sometimes overbook flights. Suppose that for a plane with 40 seats, 43 passengers have tickets. Define the random variable X as the number of ticketed passengers who show up for the flight. The probability mass function of X is given in the table below: a) What is the probability that the flight can accommodate all the ticketed passengers who show up? b) If you arc the first person on the standby list, (which means you will be the first one to get on the plane if there are any seats available after all ticketed passengers have been accommodated), what is the probability that you will be able to take the flight? c) What is the expected value of X? d) What is the variance of X? A certain telephone number is used to receive both voice calls and fax messages. Suppose that in general 80% of incoming calls do not involve fax messages. Let W be the number of the incoming calls that involve fax messages out of the next 15 calls. a) Determine the parameters of this binomial random variable (W). b) Find the probability that exactly 3 of the calls involve a fax message? Round to 3 decimals. c) What is the expected value of the number of calls among the 15 to involve a fax message? d) What is the variance of the number of calls among the 15 calls that involve a fax message? e) What is the probability that at most 5 of the calls involve a fax message? Round to 3 decimals. Let X be a random variable with the following probability distribution function: f(x) = {kx^3 0 lessthanorequalto x lessthanorequalto 4 0 otherwise a) Find the value of k. b) Find E(X). c) Find the cumulative distribution unction "F(x)" for the above pdf. d) Find P(2 lessthanorequalto X lessthanorequalto 5). e) Find the 60^th percentile of X. Round 3 decimals if necessary.

Explanation / Answer

(a) P( that flight can accomodate all the ticket passenger who show up) = P(X<=40) that means out of 43, 40 people show up for theflight = 0.04 + 0.18 + 0.22 + 0.17 + 0.24 = 0.85

(b) That question means i am the 44th person and out of first 43th, less then 40 seats have been filled up or say there is atleast one seat avaiilable = 0.04 + 0.18 + 0.22 + 0.17 = 0.61

(c) Mean or E(X) = 0.04* 36 + 0.18 * 37 + 0.22* 38 + 0.17* 39 + 0.24* 40 + 0.05 * 41+ 0.06*42 + 0.04* 43 =

= 38.98 or say 39

(d) Standerd deviation = 0.04* (36 - 39)2 + 0.18 * (37 - 39)2 + 0.22* (38- 39)2 + 0.17* (39- 39)2 + 0.24* (40- 39)2 + 0.05 * (41- 39)2+ 0.06*(42 - 39)2+ 0.04* (43- 39)2 = 2.92

Q>2 (a) Parameters for bionomial random variable

N = 15 and p = 1 - 0.8 =0.2

(b) P(X =3) = 15 C 3 * ( 0.2)3 (0.8)12 = 0.25

(c) Expected number of calls among the 15 E(W) = 0.2 * 15 = 3

(D) here we have to calculate P( X<=5; 15; 0.2) = 0.939 by binomial calculator

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