Chapter 3.3, Problem 20E

Airplane Speeds The airborne speeds in miles per hour of 21 planes are shown. Find the approximate values that correspond to the given percentiles by constructing a percentile graph.

Class	Frequency
366–386 387–407 408–428 429–449 450–470 471–491 492–512 513–533	4 2 3 2 1 2 3 4
	21

Source: The World Almanac and Book of Facts.

a. 9th

b. 20th

c. 45th

d. 60th

e. 75th

Using the same data, find the approximate percentile ranks of the following speeds in miles per hour (mph).

f. 380 mph

g. 425 mph

h. 455 mph

i. 505 mph

j. 525 mph

To determine

The $\text{9th}$ percentile of the given data.

Answer to Problem 20E

The $\text{9th}$ percentile of the given data is 376.

Explanation of Solution

Given info:

The class and frequency is shown below.

Class	Frequency
366-386	4
387-407	2
408-428	3
429-449	2
450-470	1
471-491	2
492-512	3
513-533	4
Total	21

Calculation:

Make a table as shown, and find the cumulative frequency of each class.

A Class

B Frequency ($f$)

C Cumulative frequency ($cf$)

Determine the cumulative frequency of given data and write in column C.

A Class	B Frequency ($f$)	C Cumulative frequency ($cf$)
366-386	4	4
387-407	2	6
408-428	3	9
429-449	2	11
450-470	1	12
471-491	2	14
492-512	3	17
513-533	4	21
Total	513

Locate the $\text{9th}$ percentile by $\frac{9\times 21}{100}=1.89$ observation. So the $9\text{th}$ percentiles group is $366-386$ containing the $\text{1}\text{.89th}$ observation.

Formula to calculate value corresponding percentiles is,

$P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Where,

• $l$ is lower limits of the class.
• $h$ is the width of class.
• $f$ is frequency of the class.
• $p$ is percentiles.
• $n$ is the total number.
• $c$ is preceding cumulative frequency.

Substitute 366 for $l$, 20 for $h$, 4 for $f$, 9 for $p$, 21 for $n$, and 0 for $c$.

$\begin{array}{c}{P}_9=366+\frac{20}{4}\left(\frac{9\times 21}{100}-0\right)\\ =366+\frac{20}{4}\left(1.89-0\right)\\ =366+5\left(1.89\right)\\ =366+9.45\end{array}$

Simplify the equation.

$\begin{array}{c}{P}_9=366+9.45\\ =375.45\\ \approx 376\end{array}$

Therefore, the $\text{9th}$ percentile of the data is 376.

To determine

The $\text{20th}$ percentile of the given data.

Answer to Problem 20E

The $\text{20th}$ percentile of the given data is 389.

Explanation of Solution

Calculation:

Determine the cumulative frequency of given data and write in column C.

A Class	B Frequency ($f$)	C Cumulative frequency ($cf$)
366-386	4	4
387-407	2	6
408-428	3	9
429-449	2	11
450-470	1	12
471-491	2	14
492-512	3	17
513-533	4	21
${\displaystyle \sum }$	21

Locate the $\text{20th}$ percentile by $\frac{20\times 21}{100}=4.2$ observation. So, the $\text{20th}$ percentiles group is $387-407$ containing the $\text{4}\text{.2th}$ observation.

Formula to calculate value corresponding percentiles is,

$P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Where,

• $l$ is lower limits of the class.
• $h$ is the width of class.
• $f$ is frequency of the class.
• $p$ is percentiles.
• $n$ is the total number.
• $c$ is preceding cumulative frequency.

Substitute 387 for $l$, 20 for $h$, 2 for $f$ , 20 for $p$, 21 for $n$, and 4 for $c$,

$\begin{array}{c}{P}_{20}=387+\frac{20}{2}\left(\frac{20\times 21}{100}-4\right)\\ =387+\frac{20}{2}\left(4.2-4\right)\\ =387+10\left(0.2\right)\\ =387+2\\ =389\end{array}$

Therefore, the $\text{20th}$ percentile of the data is 389.

To determine

The $\text{45th}$ percentile of the given data.

Answer to Problem 20E

The $\text{45th}$ percentile of the given data is 434.

Explanation of Solution

Locate the $\text{45th}$ percentile by $\frac{45\times 21}{100}=9.45$ observation. So, the $\text{45th}$ percentile group is $429-449$ containing the $9.45$ observation.

Formula to calculate value corresponding percentile is, $P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Substitute 429 for $l$, 20 for $h$, 2 for $f$, 45 for $p$, 21 for $n$, and 9 for $c$.

$\begin{array}{c}{P}_{45}=429+\frac{20}{2}\left(\frac{45\times 21}{100}-9\right)\\ =429+\frac{20}{2}\left(9.45-9\right)\\ =429+10\left(0.45\right)\\ =429+4.5\end{array}$

Simplify the equation,

$\begin{array}{c}{P}_{45}=429+4.5\\ =433.5\\ \approx 434\end{array}$

Therefore, the $\text{45th}$ percentile of the data is 434.

To determine

The $\text{60th}$ percentile of the given data.

Answer to Problem 20E

The $\text{60th}$ percentile of the given data is 477.

Explanation of Solution

Calculation:

Determine the cumulative frequency of given data and write in column C.

A Class	B Frequency ($f$)	C Cumulative Frequency ($cf$)
366-386	4	4
387-407	2	6
408-428	3	9
429-449	2	11
450-470	1	12
471-491	2	14
492-512	3	17
513-533	4	21
${\displaystyle \sum }$	21

Locate the $\text{60th}$ percentile by $\frac{60\times 21}{100}=12.6$ observation. So, the $\text{60th}$ percentiles group is $471-491$ containing the $12.6$ observation.

Formula to calculate value corresponding percentiles is, $P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Substitute 471 for $l$, 20 for $h$, 2 for $f$, 60 for $p$, 21 for $n$, and 12 for $c$.

$\begin{array}{c}{P}_{60}=471+\frac{20}{2}\left(\frac{60\times 21}{100}-12\right)\\ =471+\frac{20}{2}\left(12.6-12\right)\\ =471+10\left(0.6\right)\\ =471+6\end{array}$

Simplify the equation.

$\begin{array}{c}{P}_{60}=471+6\\ =477\end{array}$

Therefore, the $\text{60th}$ percentile of the data is 477.

To determine

The $\text{75th}$ percentile of the given data.

Answer to Problem 20E

The $\text{75th}$ percentile of the given data is 504.

Explanation of Solution

Calculation:

Determine the cumulative frequency of given data and write in column C.

A Class	B Frequency ($f$)	C Cumulative Frequency ($cf$)
366-386	4	4
387-407	2	6
408-428	3	9
429-449	2	11
450-470	1	12
471-491	2	14
492-512	3	17
513-533	4	21
${\displaystyle \sum }$	21

Locate the $\text{75th}$ percentile by $\frac{75\times 21}{100}=15.75$ observation. So the $\text{75th}$ percentiles group is $492-512$ containing the $15.75\text{th}$ observation.

Formula to calculate value corresponding percentiles is,

$P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Where,

• $l$ is lower limits of the class.
• $h$ is the width of class.
• $f$ is frequency of the class.
• $p$ is percentiles.
• $n$ is the total number.
• $c$ is preceding cumulative frequency.

Substitute 492 for $l$, 20 for $h$, 3 for $f$, 75 for $p$, 21 for $n$, and 14 for $c$.

$\begin{array}{c}{P}_{75}=492+\frac{20}{3}\left(\frac{75\times 21}{100}-14\right)\\ =492+\frac{20}{3}\left(15.75-14\right)\\ =492+6.67\times \left(1.75\right)\\ =492+11.67\end{array}$

Simplify the equation.

$\begin{array}{c}{P}_{75}=492+11.67\\ =503.67\\ \approx 504\end{array}$

Therefore, the $\text{75th}$ percentile of the data is 504.

To determine

The percentile rank corresponding to value 380 mph.

Answer to Problem 20E

The $\text{14th}$ percentile rank corresponding to the value is 380 mph.

Explanation of Solution

Calculation:

Determine the cumulative frequency of given data and write in column C.

A Class	B Frequency ($f$)	C Cumulative Frequency ($cf$)
366-386	4	4
387-407	2	6
408-428	3	9
429-449	2	11
450-470	1	12
471-491	2	14
492-512	3	17
513-533	4	21
${\displaystyle \sum }$	21

The speed 380 mph lies in the interval 366-386.

Formula to calculate value corresponding percentiles,

$P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Where,

• $l$ is lower limits of the class.
• $h$ is the width of class.
• $f$ is frequency of the class.
• $p$ is percentiles rank.
• $n$ is the total number.
• $c$ is preceding cumulative frequency.

Substitute 366 for $l$, 20 for $h$, 4 for $f$, 380 for $P$, 21 for $n$, and 0 for $c$.

$\begin{array}{c}380=366+\frac{20}{4}\left(\frac{p\times 21}{100}-0\right)\\ =366+\frac{20}{4}\left(0.21p-0\right)\\ =366+5\left(0.21p-0\right)\\ =366+1.05p-0\end{array}$

Simplify the equation,

$\begin{array}{c}1.05p=380-366\\ p=\frac{14}{1.05}\\ \approx 13.34\\ \approx 14\left(\text{rounding}\text{up}\right)\end{array}$

Therefore, the percentile rank for the speed 380 mph is $\text{14th}$ percentile.

To determine

The percentile rank corresponding to value 425 mph.

Answer to Problem 20E

$\text{41th}$ percentile rank corresponding to the value 425 mph.

Explanation of Solution

Calculation:

Determine the cumulative frequency of given data and write in column C.

A Class	B Frequency ($f$)	C Cumulative Frequency ($cf$)
366-386	4	4
387-407	2	6
408-428	3	9
429-449	2	11
450-470	1	12
471-491	2	14
492-512	3	17
513-533	4	21
${\displaystyle \sum }$	21

The speed 425 mph lies in the interval 408-428.

Formula to calculate value corresponding percentile is,

$P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Where,

• $l$ is lower limits of the class.
• $h$ is the width of class.
• $f$ is frequency of the class.
• $p$ is percentiles rank.
• $n$ is the total number.
• $c$ is preceding cumulative frequency.

Substitute 408 for $l$, 20 for $h$, 3 for $f$, 425 for $P$, 21 for $n$, and 6 for $c$.

$\begin{array}{c}425=408+\frac{20}{3}\left(\frac{p\times 21}{100}-6\right)\\ =408+\frac{20}{3}\left(0.21p-6\right)\\ =408+6.67\times \left(0.21p-6\right)\\ =408+1.4p-40.02\end{array}$

Simplify the equation.

$\begin{array}{c}1.4p=425-408+40.2\\ p=\frac{57.2}{1.4}\\ \approx 40.85\\ \approx 41\left(\text{rounding}\text{up}\right)\end{array}$

Therefore, the percentile rank for the speed 425mph is $\text{41th}$ percentile.

To determine

The percentile rank corresponding to value 455.

Answer to Problem 20E

$\text{54th}$ percentile rank corresponding to the value 455.

Explanation of Solution

Calculation:

Determine the cumulative frequency of given data and write in column C.

A Class	B Frequency ($f$)	C Cumulative Frequency ($cf$)
366-386	4	4
387-407	2	6
408-428	3	9
429-449	2	11
450-470	1	12
471-491	2	14
492-512	3	17
513-533	4	21
${\displaystyle \sum }$	21

The speed 455 mph lies in the interval 450-470.

Formula to calculate value corresponding percentile is,

$P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Where,

• $l$ is lower limits of the class.
• $h$ is the width of class.
• $f$ is frequency of the class.
• $p$ is percentiles rank.
• $n$ is the total number.
• $c$ is preceding cumulative frequency.

Substitute 450 for $l$, 20 for $h$, 1 for $f$, 455 for $P$, 21 for $n$, and 11 for $c$.

$\begin{array}{c}455=450+\frac{20}{1}\left(\frac{p\times 21}{100}-11\right)\\ =450+\frac{20}{1}\left(0.21p-11\right)\\ =450+20\times \left(0.21p-11\right)\\ =450+4.2p-220\end{array}$

Simplify the equation,

$\begin{array}{c}4.2p=455-450+220\\ p=\frac{225}{4.2}\\ \approx 53.57\\ \approx 54\left(\text{rounding}\text{up}\right)\end{array}$

Therefore, the percentile rank for the speed 455 mph is $\text{54th}$ percentile.

To determine

The percentile rank corresponding to value 505.

Answer to Problem 20E

The percentile rank corresponding to value 505 is $\text{76th}$ percentile.

Explanation of Solution

Calculation:

Determine the cumulative frequency of given data and write in column C.

A Class	B Frequency ($f$)	C Cumulative Frequency ($cf$)
366-386	4	4
387-407	2	6
408-428	3	9
429-449	2	11
450-470	1	12
471-491	2	14
492-512	3	17
513-533	4	21
${\displaystyle \sum }$	21

The speed 505mph lies in the interval 492-512.

Formula to calculate value corresponding percentile is,

$P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Where,

• $l$ is lower limits of the class.
• $h$ is the width of class.
• $f$ is frequency of the class.
• $p$ is percentiles rank.
• $n$ is the total number.
• $c$ is preceding cumulative frequency.

Substitute 492 for $l$, 20 for $h$, 3 for $f$, 505 for $P$, 21 for $n$, and 14 for $c$.

$\begin{array}{c}505=492+\frac{20}{3}\left(\frac{p\times 21}{100}-14\right)\\ =492+\frac{20}{3}\left(0.21p-14\right)\\ =492+6.67\times \left(0.21p-14\right)\\ =492+1.4p-93.38\end{array}$

Simplify the equation,

$\begin{array}{c}1.4p=505-492+93.38\\ p=\frac{106.38}{1.4}\\ \approx 75.98\\ \approx 76\left(\text{rounding}\text{up}\right)\end{array}$

Therefore, the percentile rank for the speed 505 mph is $\text{76th}$ percentiles.

To determine

The percentile rank corresponding to value 525.

Answer to Problem 20E

The percentile rank for the speed 525 mph is $\text{93th}$ percentile.

Explanation of Solution

Calculation:

Determine the cumulative frequency of given data and write in column C.

A Class	B Frequency ($f$)	C Cumulative Frequency ($cf$)
366-386	4	4
387-407	2	6
408-428	3	9
429-449	2	11
450-470	1	12
471-491	2	14
492-512	3	17
513-533	4	21
${\displaystyle \sum }$	21

The speed 525 mph lies in the interval 513-533.

Formula to calculate value corresponding percentiles is,

$P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)$

Where,

• $l$ is lower limits of the class.
• $h$ is the width of class.
• $f$ is frequency of the class.
• $p$ is percentiles rank.
• $n$ is the total number.
• $c$ is preceding cumulative frequency.

Substitute 513 for $l$, 20 for $h$, 4 for $f$, 525 for $P$, 21 for $n$, and 17 for $c$.

$\begin{array}{c}525=513+\frac{20}{4}\left(\frac{p\times 21}{100}-17\right)\\ =513+\frac{20}{4}\left(0.21p-17\right)\\ =513+5\left(0.21p-17\right)\\ =513+1.05p-85\end{array}$

Simplify the equation,

$\begin{array}{c}1.05p=525-513+85\\ p=\frac{97}{1.05}\\ \approx 92.38\\ \approx 93\left(\text{rounding}\text{up}\right)\end{array}$

Therefore, the percentile rank for the speed 525 mph is $\text{93th}$ percentile.