Chapter 6.5, Problem 22P

(a)

To determine

Find the probability that total taxi and take off time would be less than 320 minutes for 36 jets on a given runway.

Answer to Problem 22P

The probability that total taxi and take off time would be less than 320 minutes for 36 jets on a given runway is 0.8238.

Explanation of Solution

Calculation:

Central limit theorem:

The characteristics of the sampling distribution $\overline{x}$, for all the sample size n is determined using the central limit theorem. Based on central limit theorem the sample size n must be greater than or equal to 30 for making the distribution of sample means $\overline{x}$ to be approximately normal, even if the population is not normally distributed.

Sampling distribution:

For x distribution the sampling distribution of $\overline{x}$ is,

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\end{array}$

In the formula, $\mu$ denotes population mean, $\sigma$ denotes population standard deviation, n denotes the sample size.

The $\overline{x}$ can be converted into standard normal distribution as,

$z=\frac{\overline{x}-{\mu }_{\overline{x}}}{{\sigma }_{\overline{x}}}$

In the formula, $\overline{x}$ is the mean raw score, ${\mu }_{\overline{x}}$ is the mean of the sampling distribution and ${\sigma }_{\overline{x}}$ is the standard deviation of the sampling distribution.

The random variable x is defined as taxi and take-off time for commercial jets. The $x_i$ is denotes the take-off time for $i^{\text{th}}$ commercial jets $\left(i=1,2,\mathrm{...},36\right)$. The sum of the take-off time for all the 36 commercial jets is, $w=x_1+x_2+\mathrm{...}+x_{36}$. The total taxi and take-off time for 36 commercial jets on runway (w) is less than 320 minutes has to be calculated which is denoted as $P\left(w<320\right)$.

Consider the expression $w<320$. Divide both sides of the expression with 36.

$\begin{array}{c}w<320\\ \frac{w}{36}<\frac{320}{36}\\ \frac{w}{36}<8.889\end{array}$

The sum of taxi and take off time for 36 commercial jets on runway is denoted by w. That is,

$\begin{array}{c}\frac{w}{36}=\frac{x_1+x_2+\mathrm{...}+x_{36}}{36}\\ =\overline{x}\end{array}$

The total taxi and take off time for 36 commercial jets on runway (w) is less than 320 can be written as,

$\begin{array}{c}P\left(w<320\right)=P\left(\frac{w}{36}<8.889\right)\\ =P\left(\overline{x}<8.889\right)\end{array}$

Hence, the quantities $P\left(w<320\right)$ and $P\left(\overline{x}<8.889\right)$ are equal.

The number of commercial jets on runway in the study is $n=36$. The sample size is greater than 36; based on the central limit theorem the distribution of sample means $\overline{x}$ is approximately normal.

The mean of the x distributions is $\mu =8.5\text{minutes}$. The mean of sampling distribution is equal to the mean of population distribution. That is,

$\begin{array}{c}{\mu }_{\overline{x}}=\mu \\ =8.5\end{array}$

The standard deviation of the x distributions is $\sigma =2.5\text{minutes}$. The standard deviation of $\overline{x}$ distribution is,

$\begin{array}{c}{\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}}\\ =\frac{2.5}{\sqrt{36}}\\ =\frac{2.5}{6}\\ =0.4167\end{array}$

Z score:

Substitute $\overline{x}$ as 8.889, ${\mu }_{\overline{x}}$ as 8.5 and ${\sigma }_{\overline{x}}$ as 0.4167 in the formula of z score.

$\begin{array}{c}z=\frac{8.889-8.5}{0.4167}\\ =\frac{0.389}{0.4167}\\ =0.93\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than 0.93.

• Locate the value 0.9 in column z.
• Locate the value 0.03 in top row.
• The intersecting value of row and column is 0.8238.

The probability is,

$\begin{array}{c}P\left(\overline{x}<8.889\right)=P\left(z<0.93\right)\\ =0.8238\end{array}$

Hence, the probability that total taxi and take off time would be less than 320 minutes for 36 jets on a given runway is 0.8238.

(b)

To determine

Find the probability that total taxi and take off time would be more than 275 minutes for 36 jets on a given runway.

Answer to Problem 22P

The probability that total taxi and take off time would be more than 275 minutes for 36 jets on a given runway is 0.9808.

Explanation of Solution

Calculation:

The total taxi and take-off time for 36 commercial jets on runway (w) is more than 275 minutes has to be calculated which is denoted as $P\left(w>275\right)$.

Consider the expression $w>275$. Divide both sides of the expression with 36.

$\begin{array}{c}w>275\\ \frac{w}{36}>\frac{275}{36}\\ \frac{w}{36}>7.639\end{array}$

The sum of taxi and take off time for 36 commercial jets on runway is denoted by w. That is,

$\begin{array}{c}\frac{w}{36}=\frac{x_1+x_2+\mathrm{...}+x_{36}}{36}\\ =\overline{x}\end{array}$

The total taxi and take off time for 36 commercial jets on runway (w) is more than 275 can be written as,

$\begin{array}{c}P\left(w>275\right)=P\left(\frac{w}{36}>7.639\right)\\ =P\left(\overline{x}>7.639\right)\end{array}$

Hence, the quantities $P\left(w>275\right)$ and $P\left(\overline{x}>7.639\right)$ are equal.

Z score:

Substitute $\overline{x}$ as 7.639, ${\mu }_{\overline{x}}$ as 8.5 and ${\sigma }_{\overline{x}}$ as 0.4167 in the formula of z score.

$\begin{array}{c}z=\frac{7.639-8.5}{0.4167}\\ =\frac{-0.861}{0.4167}\\ =-2.07\end{array}$

Use the Appendix II: Tables, Table 5: Areas of a Standard Normal Distribution: to obtain probability less than –2.07.

• Locate the value –2.0 in column z.
• Locate the value 0.07 in top row.
• The intersecting value of row and column is 0.0192.

The probability is,

$\begin{array}{c}P\left(\overline{x}>7.639\right)=P\left(z>-2.07\right)\\ =1-P\left(z\le -2.07\right)\\ =1-0.0192\\ =0.9808\end{array}$

Hence, the probability that total taxi and take off time would be more than 275 minutes for 36 jets on a given runway is 0.9808.

(c)

To determine

Find the probability that total taxi and take off time would be between 275 minutes and 320 minutes for 36 jets on a given runway.

Answer to Problem 22P

The probability that total taxi and take off time would be between 275 minutes and 320 minutes for 36 jets on a given runway is 0.8046.

Explanation of Solution

Calculation:

From part (a), the quantity 320 minutes is equal to 8.889 after conversion. The z score is 0.93. The probability value for less than 0.93 from normal table is 0.8238.

From part (b), the quantity 275 minutes is equal to 7.639 after conversion. The z score is $–2.07$. The probability value for less than –2.07 from normal table is 0.0192.

The probability is,

$\begin{array}{c}P\left(275<w<320\right)=P\left(7.639<\overline{x}<8.889\right)\\ =P\left(-2.07<z<0.93\right)\\ =P\left(z<0.93\right)-P\left(z<-2.07\right)\\ =0.8238-0.0192\\ =0.8046\end{array}$

Hence, the probability that total taxi and take off time would be between 275 minutes and 320 minutes for 36 jets on a given runway is 0.8046.