Suppose that the probability that a passenger will miss a flight is 0.0959. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must 53 passengers. (a) If 55 tickets are sold, what is the probability that 54 or 55 passengers show up for the flight resulting in an overbooked flight? (b) Suppose that 59 tickets are sold. What is the probability that a passenger will have to be "bumped"? (c) For a plane with seating capacity of 250 passengers, what is the largest number of tickets that can be sold to keep the probability of a passenger being "bumped" below 1%? (a) The probability of an overbooked flight is (Round to four decimal places as needed.) (b) The probability that a passenger will have to be bumped is (Round to four decimal places as needed.) (c) For a plane with seating capacity of 250 passengers, the largest number of tickets that can be sold while keeping the probability of a passenger being "bumped" below 1% is. (Type a whole number.)

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Suppose that the probability that a passenger will miss a flight is 0.0959. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must be "bumped" from the flight. Suppose that an airplane has a seating capacity of
53 passengers.
(a) If 55 tickets are sold, what is the probability that 54 or 55 passengers show up for the flight resulting in an overbooked flight?
(b) Suppose that 59 tickets are sold. What is the probability that a passenger will have to be "bumped"?
(c) For a plane with seating capacity of 250 passengers, what is the largest number of tickets that can be sold to keep the probability of a passenger being "bumped" below 1%?
(a) The probability of an overbooked flight is
(Round to four decimal places as needed.)
(b) The probability that a passenger will have to be bumped is
(Round to four decimal places as needed.)
C
(c) For a plane with seating capacity of 250 passengers, the largest number of tickets that can be sold while keeping the probability of a passenger being "bumped" below 1% is.
(Type a whole number.)
Transcribed Image Text:Suppose that the probability that a passenger will miss a flight is 0.0959. Airlines do not like flights with empty seats, but it is also not desirable to have overbooked flights because passengers must be "bumped" from the flight. Suppose that an airplane has a seating capacity of 53 passengers. (a) If 55 tickets are sold, what is the probability that 54 or 55 passengers show up for the flight resulting in an overbooked flight? (b) Suppose that 59 tickets are sold. What is the probability that a passenger will have to be "bumped"? (c) For a plane with seating capacity of 250 passengers, what is the largest number of tickets that can be sold to keep the probability of a passenger being "bumped" below 1%? (a) The probability of an overbooked flight is (Round to four decimal places as needed.) (b) The probability that a passenger will have to be bumped is (Round to four decimal places as needed.) C (c) For a plane with seating capacity of 250 passengers, the largest number of tickets that can be sold while keeping the probability of a passenger being "bumped" below 1% is. (Type a whole number.)
Use n = 6 and p = 0.7 to complete parts (a) through (d) below.
(a) Construct a binomial probability distribution with the given parameters.
P(x)
X
0
1
2
3
4
5
6
(Round to four decimal places as needed.)
(b) Compute the mean and standard deviation of the random variable using μx = [x• P(x)] and ox=
Hx=
(Round to two decimal places as needed.)
6x =
(Round to two decimal places as needed.)
(c) Compute the mean and standard deviation, using μx = np and ox=√√np(1-p).
(Round to two decimal places as needed.)
(Round to two decimal places as needed.)
Hx =
0x =
(d) Draw a graph of the probability distribution and comment on its shape.
Which graph below shows the probability distribution?
OA.
AP(x)
0.5-
0.25-
The binomial probability distribution is
O B.
AP(x)
0.5-
0.25-
√Σ[x² • P(x)]-H².
C
O C.
AP(x)
0.5-
0.25-
3
Q
O D.
AP(x)
0.5-
0.25-
Transcribed Image Text:Use n = 6 and p = 0.7 to complete parts (a) through (d) below. (a) Construct a binomial probability distribution with the given parameters. P(x) X 0 1 2 3 4 5 6 (Round to four decimal places as needed.) (b) Compute the mean and standard deviation of the random variable using μx = [x• P(x)] and ox= Hx= (Round to two decimal places as needed.) 6x = (Round to two decimal places as needed.) (c) Compute the mean and standard deviation, using μx = np and ox=√√np(1-p). (Round to two decimal places as needed.) (Round to two decimal places as needed.) Hx = 0x = (d) Draw a graph of the probability distribution and comment on its shape. Which graph below shows the probability distribution? OA. AP(x) 0.5- 0.25- The binomial probability distribution is O B. AP(x) 0.5- 0.25- √Σ[x² • P(x)]-H². C O C. AP(x) 0.5- 0.25- 3 Q O D. AP(x) 0.5- 0.25-
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Hello! As you have posted 2 different questions, we are answering the first question. In case you require the unanswered question also, kindly re-post them as separate question.

From the given information,

The probability that a passenger will miss a flight is 0.0959.

The probability that the passenger will not miss the flight is 1-0.0959=0.9041.

a)

Let X be the number of passengers show up for the flight follows Binomial distribution with n=55 and p=0.9041.

PX=x=nxpx1-pn-x               =55x0.9041x1-0.904155-x

The probability that 54 or 55 passengers show up for the flight resulting in an overbooked flight is,

PX=54+PX=55=55540.9041541-0.904155-54+55550.9041551-0.904155-55                                      =0.0228+0.0039                                      =0.0267

 

 

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