A small commuter airline is concerned about reservation no s

A small commuter airline is concerned about reservation \"no shows\" and correspondingly, how much they should overbook flights to compensate. Assume their commuter planes will hold 15 people. Industry research indicates that 20% o f the people making a reservation will not show up for a flight. Whether or not one person takes the flight is considered to be independent of other persons holding reservations. What probability model would be appropriate for the number of passengers that actually take the flight? If the airlines decide to book 18 people for each flight, how often will there be at least one person who will not get a seat? Seven plants are operated by a garment manufacturer. They feel there is a ten percent chance for a strike at any one plant and the risk o f a strike at one plant is independent of the risk of a strike at another plant. Let X - number of plants of the garment manufacturer that strike. Determine the probability distribution for X. Interpret the results for P (X = 0), P(X = 4). and P(X = 7). Compute the expected value for X. Compute the standard deviation for X. Is this value large in relation to the expected value? In what units is the standard deviation expressed?

Solution

1. let X be the number of persons who could actually took the flight.

the probability that a randomly chosen person could take the flight=1-probability that he could not take the flight=1-0.2=0.8

the commuter planes hold 15 persons. moreover the bookings of each person is independent of the other.

a) so the appropriate model for X is binomial

X~Bin(15,0.8)

pmf of X is f(x)=¹⁵C_x*0.8^x*0.2^15-x          x=0,1,2,.........,15

b) airlines decided to book for 18 people.there were seat for 15 people

so P[at least one will not get a seat]=1-P[everyone will get seat]=1-P[out of 18 people , 3 will not show up.hence the other 15 will get seat]=1-0.2*0.2*0.2=0.992 [answer]

2) X: number of plants of the garment manufacturer that strike.

there are seven plants over all and the probability of strike at ony one plant is 0.1

moreover the strikes are independent.

a) the probability distribution of X is Bin(7,0.1)

pmf of X is f(x)=⁷C_x*0.1^x*0.9^7-x              x=0,1,2,...,7

b) P[X=0]= 0.478297     so there is 47.8297% chance that no plants of the garment manufacturer that strikes.

    P[X=4]= 0.002552     so there is 0.2552% chance that 4 plants of the garment manufacturer that strikes.

    P[X=7]=0.000000      so there is no chance that all the seven plants of the garment manufacturer that strikes.

c) expected value of X is E[X]=7*0.1=0.7 [answer]

d)standard deviation=sqrt(7*0.1*0.9)=0.7937 [answer]

yes, it has a large value compared to expected value.

here standard deviation is measured in the same unit as that of X.