What Is the Probability of Having Enough Seats on an Overbooked Flight?

[subopolois]

Homework Statement

a airline finds that 5% of people who bought a plane ticket do not show up for the flight. if the airline sells 160 tickets for a flight that has 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?

Homework Equations

central limit theorem

The Attempt at a Solution

since 5% of people don't show and the airline sells 160 seats:
160x0.05= 8
so of the 160 that buy a ticket 8 people will not show up:
160-8= 152
since there are 155 seats and 152 show up for the flight:
152/155= 0.9806

is this right? I am really not sure if what i did war right or not, if it isnt, can someone guide me in the right direction?

 

[HallsofIvy]

No, that's not right. "8" is the expected value of the number of people who do not show up, not a probability. 152 is the expected value of the number of people who do show up but dividing by the number of seats still does not give a probability.

This is a binomial distribution with n= 160, p= 0.95 and q= 0.05. What you want to find is the probability that the number of people who do not show up is greater than or equal to 5. The probability that NONE fail to show up is
$$\left(\begin{array}{c}155 \\ 0\end{array}\right)(0.05)^0(0.95)^{155}$$.
The probability that exactly one fails to show up is
$$\left(\begin{array}{c}155 \\ 1\end{array}\right)(0.05)^1(0.95)^{154}$$
etc.
Find the probability that 0, 1, 2, 3, 4 fail to show up and add. That will be the probability that 4 or fewer fail to show up. The probability that 5 or more fail to show is 1 minus that.

Another way to do this, since 155 is such a large number, is to use the normal distribution approximation to the binomial distribution. Since the normal distribution allows real number variables, use the "half-integer" correction: any thing less than 5.5 rounds to 5- use the normal distribution with mean (0.5)(155), standard distribution $\sqrt{(0.5)(0.95)(155)}$ and find the probability that the number of "no-shows" is greater than 5.5

 

[subopolois]

No, that's not right. "8" is the expected value of the number of people who do not show up, not a probability. 152 is the expected value of the number of people who do show up but dividing by the number of seats still does not give a probability.

This is a binomial distribution with n= 160, p= 0.95 and q= 0.05. What you want to find is the probability that the number of people who do not show up is greater than or equal to 5. The probability that NONE fail to show up is
$$\left(\begin{array}{c}155 \\ 0\end{array}\right)(0.05)^0(0.95)^{155}$$.
The probability that exactly one fails to show up is
$$\left(\begin{array}{c}155 \\ 1\end{array}\right)(0.05)^1(0.95)^{154}$$
etc.
Find the probability that 0, 1, 2, 3, 4 fail to show up and add. That will be the probability that 4 or fewer fail to show up. The probability that 5 or more fail to show is 1 minus that.

Another way to do this, since 155 is such a large number, is to use the normal distribution approximation to the binomial distribution. Since the normal distribution allows real number variables, use the "half-integer" correction: any thing less than 5.5 rounds to 5- use the normal distribution with mean (0.5)(155), standard distribution $\sqrt{(0.5)(0.95)(155)}$ and find the probability that the number of "no-shows" is greater than 5.5

just double checking, for the second method, is the mean (0.05)(155)? not (0.5)(155)? or is it just a typo?

 

[subopolois]

No, that's not right. "8" is the expected value of the number of people who do not show up, not a probability. 152 is the expected value of the number of people who do show up but dividing by the number of seats still does not give a probability.

This is a binomial distribution with n= 160, p= 0.95 and q= 0.05. What you want to find is the probability that the number of people who do not show up is greater than or equal to 5. The probability that NONE fail to show up is
$$\left(\begin{array}{c}155 \\ 0\end{array}\right)(0.05)^0(0.95)^{155}$$.
The probability that exactly one fails to show up is
$$\left(\begin{array}{c}155 \\ 1\end{array}\right)(0.05)^1(0.95)^{154}$$
etc.
Find the probability that 0, 1, 2, 3, 4 fail to show up and add. That will be the probability that 4 or fewer fail to show up. The probability that 5 or more fail to show is 1 minus that.

Another way to do this, since 155 is such a large number, is to use the normal distribution approximation to the binomial distribution. Since the normal distribution allows real number variables, use the "half-integer" correction: any thing less than 5.5 rounds to 5- use the normal distribution with mean (0.5)(155), standard distribution $\sqrt{(0.5)(0.95)(155)}$ and find the probability that the number of "no-shows" is greater than 5.5

alright, i understand the first part, and somewhat the second, here is what i have done:
n=155; mean= (0.05)(155); SD= 2.71
P(x>5.5)= (x>5.5-7.75/2.71(sqrt155)
= (z>-2.25/0.2178)
= (z>-10.14)
is this right? because that z-score seems way to high, or is it something i did?