Chapter 3, Problem 3.3.23RE

The number of police calls a small police department received each month is shown in the frequency distribution.

Class limits	Frequency
39.9–42.8	2
42.9–45.8	2
45.9–48.8	5
48.9–51.8	5
51.9–54.8	12
54.9–57.8	3

a. Construct a percentile graph.

b. Find the values that correspond to the 35th, 65th, and 85th percentiles.

c. Find the percentile of values 44, 48, and 54.

To determine

To construct: A percentile graph for the given data.

Answer to Problem 3.3.23RE

Percentile graph is shown below.

Explanation of Solution

Given info:

Class limit	Frequency
39.9-42.8	2
42.8-45.8	2
45.9-48.8	5
48.9-51.8	5
51.9-54.8	12
54.9-57.8	3

Calculation:

Make a table for calculating percentile graph is shown below,

A Class limit	B Frequency	C Cumulative frequency	D Cumulative percent

Determine cumulative frequency and place in column C.

Determine the cumulative percentage and place in column D.

$\text{Cumulative}\%=\frac{\text{cumulative}\text{frequency}}{n}\times 100\%$

Put 2 for cumulative frequency and 29 for $n$ in the above formula to calculate the cumulative percentage for first class.

$\begin{array}{c}\text{Cumulative}\%=\frac{2}{29}\times 100\%\\ =6.89\%\end{array}$

Put 4 for cumulative frequency and 29 for $n$ in the above formula to calculate the cumulative percentage for second class.

$\begin{array}{c}\text{Cumulative}\%=\frac{4}{29}\times 100\%\\ =13.79\%\end{array}$

Put 9 for cumulative frequency and 29 for $n$ in the above formula to calculate the cumulative percentage for third class.

$\begin{array}{c}\text{Cumulative}\%=\frac{9}{29}\times 100\%\\ =31.03\%\end{array}$

Put 14 for cumulative frequency and 29 for $n$ in the above formula to calculate the cumulative percentage for fourth class.

$\begin{array}{c}\text{Cumulative}\%=\frac{14}{29}\times 100\%\\ =73.68\%\end{array}$

Put 26 for cumulative frequency and 29 for $n$ in the above formula to calculate the cumulative percentage for fifth class.

$\begin{array}{c}\text{Cumulative}\%=\frac{26}{29}\times 100\%\\ =89.65\%\end{array}$

Thus, the table for calculating percentile graph is,

A Class limit	B Frequency	C Cumulative frequency	D Cumulative percent
39.9-42.8	2	2	6.89
42.9-45.8	2	4	13.79
45.9-48.8	5	9	31.03
48.9-51.8	5	14	73.68
51.9-54.8	12	26	89.65
54.9-57.8	3	29	100

Graph the data using class boundaries for the x axis and the percentage for the y axis. Plot the cumulative percent and join the lines. Thus, the percentile graph is obtained as shown below.

To determine

The values that correspond to the 35^th, 65^th and 85^th percentiles.

Answer to Problem 3.3.23RE

The percentile values corresponding to 35^th, 65^th and 85^th are 50, 53 and 55.

Explanation of Solution

Calculation:

Formula to calculate the percentile for the grouped data is given below,

$P_m=l+\frac{h}{f}\left(\frac{m.n}{100}-c\right)$

Where,

• $l$ is the lower boundary of the class of the $m^{th}$ percentile.
• $h$ is the width of the class.
• $f$ is frequency of the class.
• $n$ is the total number of the frequencies.

Table for calculating percentile value,

A Class limit	B Frequency	C Cumulative frequency	D Cumulative percent
39.9-42.8	2	2	6.89
42.9-45.8	2	4	13.79
45.9-48.8	5	9	31.03
48.9-51.8	5	14	73.68
51.9-54.8	12	26	89.65
54.9-57.8	3	29	100

Locate the 35^th percentiles by $\frac{35\times 29}{100}=10.12$. So 35^th percentiles lies in the interval 48.9 and 51.8.

Substitute 35 for m, 48.9 for $l$, 3 for $h$, 5 for $f$, 29 for $n$ and 9 for $c$ in the above formula,

$\begin{array}{c}{P}_{35}=48.9+\frac{3}{5}\left(\frac{35\times 29}{100}-9\right)\\ =48.9+\frac{3}{5}\left(10.12-9\right)\\ =48.9+0.672\\ =49.57\end{array}$

Thus, the 35^th percentile of the data is approximately 50.

Locate the 65^th percentiles by $\frac{65\times 29}{100}=18.85$. So 65^th percentiles lies in the interval 51.9 and 54.8.

Substitute 65 for m, 51.9 for $l$, 3 for $h$, 12 for $f$, 29 for $n$ and 14 for $c$ in the above formula,

$\begin{array}{c}{P}_{65}=51.9+\frac{3}{12}\left(\frac{65\times 29}{100}-14\right)\\ =51.9+\frac{3}{12}\left(18.85-14\right)\\ =51.9+1.21\\ =53.11\end{array}$

Thus, the 65^th percentile of the data is approximately 53.

Locate the 85^th percentiles by $\frac{85\times 29}{100}=24.65$. So 85^th percentiles lies in the interval 51.9 and 54.8.

Substitute 85 for m, 51.9 for $l$, 3 for $h$, 12 for $f$, 29 for $n$ and 14 for $c$ in the above formula,

$\begin{array}{c}{P}_{85}=51.9+\frac{3}{12}\left(\frac{85\times 29}{100}-14\right)\\ =51.9+\frac{3}{12}\left(24.65-14\right)\\ =51.9+2.66\\ =54.56\end{array}$

Thus, the 85^th percentile of the data is approximately 55.

To determine

The percentile rank of values 44, 48 and 54.

Answer to Problem 3.3.23RE

The percentile rank for values 44, 48 and 54 are 9^th, 26^th and 77^th respectively.

Explanation of Solution

Calculation:

Formula to calculate the percentile rank for the grouped data is given below,

$P_m=l+\frac{h}{f}\left(\frac{m.n}{100}-c\right)$

Where,

• $l$ is the lower boundary of the class of the $m^{th}$ percentile.
• $h$ is the width of the class.
• $f$ is frequency of the class.
• $n$ is the total number of the frequencies.
• $c$ is the cumulative frequency of the class immediate preceding to the class.

Value 44 lies in the interval 42.9-45.8.

Substitute 44 for $P_m$, 42.9 for $l$, 3 for $h$, 2 for $f$, 29 for $n$ and 2 for $c$ in the above formula,

$\begin{array}{c}44=42.9+\frac{3}{2}\left(\frac{m\times 29}{100}-2\right)\\ =42.9+1.5\left(0.29m-2\right)\\ =42.9+0.44m-3\\ =39.9+0.44m\end{array}$

Simplify the equation,

$\begin{array}{c}0.44m=44-39.9\\ m=\frac{4.1}{0.44}\\ =9.31\\ =9\end{array}$

Thus, the percentile rank of the value 44 is 9^th percentile.

The value 48 lies in the interval 48.9-51.8.

Substitute 48 for $P_m$, 48.9 for $l$, 3 for $h$, 5 for $f$, 29 for $n$ and 9 for $c$ in the above formula,

$\begin{array}{c}48=48.9+\frac{3}{5}\left(\frac{m\times 29}{100}-9\right)\\ =48.9+0.6\left(0.29m-9\right)\\ =48.9+0.174m-5.4\\ =43.5+0.174m\end{array}$

Simplify the equation,

$\begin{array}{c}0.174m=48-43.5\\ m=\frac{4.5}{0.174}\\ =25.86\\ =26\end{array}$

Thus, the percentile rank of the value 48 is 26^th percentile.

The value 54 lies in the interval 51.9-54.8.

Substitute 54 for $P_m$ 51.9 for $l$, 3 for $h$, 12 for $f$, 29 for $n$ and 14 for $c$ in the above formula,

$\begin{array}{c}54=51.9+\frac{3}{12}\left(\frac{m\times 29}{100}-14\right)\\ =51.9+\frac{3}{12}\left(0.29m-14\right)\\ =51.9+0.0725m-3.5\\ =48.4+0.0725m\end{array}$

Simplify the equation,

$\begin{array}{c}0.0725m=54-48.4\\ m=\frac{5.6}{0.0725}\\ =77.24\\ \approx 77\end{array}$

Thus, the percentile rank of the value 54 is 77^th percentile.