91Ó°ÊÓ

There is a 0.0995 probability that a person will not show up for the flight. There are 213 seats in the specified aircraft.

It is given that the probability that a person will not show up for the flight is equal to 0.0995.

It is required to determine the total number of reservations that could be accepted to accommodate at least 95% of the people who show up.

In terms of z-score, the value of the z-score corresponding to the probability of 0.95 is equal to 1.645.

Let X denote the number of people who do not show up after making a reservation.

The probability that a person will not show up after making a reservation is equal to p=0.0995.

Here, n=213.

The following formula is used to convert the z-score to the required sample value:

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ x=\mathrm{μ}+z\mathrm{σ} \end{gathered}$$

The mean value is computed below:

$$\begin{gathered} \mathrm{μ}=np \\ =213×0.0995 \\ =21.1935 \end{gathered}$$

The standard deviation is computed below:

$$\begin{gathered} \mathrm{σ}=\sqrt{npq} \\ =\sqrt{np(1-p)} \\ =\sqrt{213×0.0995×(1-0.0995)} \\ ≈4.3686 \end{gathered}$$

Thus, the value of x is equal to:

$$\begin{gathered} x=\mathrm{μ}+z\mathrm{σ} \\ =21.1935+1.645×4.3686 \\ =28.37 \\ ≈28 \end{gathered}$$

Now, the total number of reservations that can be accepted, including the number of reservations that are not fulfilled, is computed below:

$213+28=241$

Therefore, 241 reservations could be accepted for the flight, keeping in mind at least a 0.95 probability that all reservation holders who show will be accommodated.