Chapter 4, Problem 1SE

An airplane has 100 seats for passengers. Assume that the probability that a person holding a ticket appears for the flight is 0.90. If the airline sells 105 tickets, what is the probability that everyone who appears for the flight will get a seat?

To determine

Find the probability that everyone who appears for the flight will get a seat, if the airline sells 105 tickets.

Answer to Problem 1SE

The probability that everyone who appears for the flight will get a seat is 0.9774.

Explanation of Solution

Given info:

There are 100 seats for passengers in an airplane. The probability that a person holding a ticket appears for the flight is 0.90.

Calculation:

The random variable X is defined as the number of people out of 105 who appear for the flight. Then the random variable X follows binomial distribution with parameters $n=105\text{ and }p=0.9$.

The condition for the normal approximation to the binomial probability distribution is $np>10\text{ and }n\left(1-p\right)>10$.

That is,

$\begin{array}{c}np=105\left(0.90\right)\\ =94.5\\ >10\end{array}$

$\begin{array}{c}n\left(1-p\right)=\left(105\right)\left(1-0.9\right)\\ =\left(105\right)\left(0.1\right)\\ =10.5\end{array}$

Thus, the normal distribution can be used to approximate the binomial distribution.

Mean:

${\mu }_X=np$

Substitute 105 for n and 0.90 for p.

$\begin{array}{c}{\mu }_X=\left(105\right)\left(0.90\right)\\ =94.5\end{array}$

Standard deviation:

${\sigma }_X=\sqrt{np(1-p)}$

Substitute n as 60 and p as 0.40.

$\begin{array}{c}{\sigma }_X=\sqrt{\left(105\right)\left(0.90\right)\left(1-0.90\right)}\\ =\sqrt{94.5}\\ =3.0741\end{array}$

The approximate binomial probability using the normal distribution is,

$P\left(X\le a\right)\approx P\left(X<a+0.5\right)$

By using continuity correction, the value 0.5 is added from the value of 100.

Substitute 100 for a,

$\begin{array}{c}P\left(X\le 100\right)=P\left(X<100+0.5\right)\\ =P\left(X<100.5\right)\end{array}$

The formula to convert X values into z score is,

$z=\frac{X-np}{\sqrt{np(1-p)}}$

Substitute 94.5 for np and 3.0741 for $\sqrt{np(1-p)}$.

$\begin{array}{c}P\left(X<100.5\right)=P\left(z\le \frac{100.5-94.5}{3.0741}\right)\\ =P\left(z\le \frac{6}{3.0741}\right)\\ =P\left(z\le 1.95\right)\end{array}$

The above probability can be obtained by finding the areas to the left of 1.95.

Use Table A.2: Cumulative Normal Distribution to find the area.

Procedure:

For z at 1.95,

• Locate 1.9 in the left column of the Table A.2.
• Obtain the value in the corresponding row below 0.05.

That is, $P\left(z\le 1.95\right)=0.9744$

Thus, the probability that everyone who appears for the flight will get a seat, if the airline sells 105 tickets is 0.9774.