Chapter 3, Problem 3.3.23RE

a.

To determine

To construct: A percentile graph for the given data.

Answer to Problem 3.3.23RE

The percentile graph for the given data is as follows,

Explanation of Solution

Given info:

The data represents the salaries (in millions of dollars) for 29 NFL teams for the 1999 2000 season.

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

Calculation:

The cumulative frequency for the distribution is calculated and tabulated below,

Class limits	Frequency	Cumulative frequency
39.9-42.8	2	2
42.9-45.8	2	$2+2=4$
45.9-48.8	5	$4+5=9$
48.9-51.8	5	$9+5=14$
51.8-54.8	12	$14+12=26$
54.9-57.8	3	$26+3=29$

The formula to calculate the cumulative percentage is as follows,

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

For the cumulative frequency 2:

Substitute cumulative frequency as 2 and n as 29 in the formula,

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

Similarly for remaining values the cumulative percent are tabulated below

Class limit	Frequency	Cumulative frequency	Cumulative percent
39.9-42.8	2	2	6.89
42.9-45.8	2	4	$\frac{4}{29}\times 100=13.79$
45.9-48.8	5	9	$\frac{9}{29}\times 100=31.03$
48.9-51.8	5	14	$\frac{14}{29}\times 100=73.68$
51.9-54.8	12	26	$\frac{26}{29}\times 100=89.65$
54.9-57.8	3	29	$\frac{29}{29}\times 100=100$
	${\displaystyle \sum f}=29$

Software procedure:

Step-by-step software procedure to draw ogive curve using MINITAB software is as follows:

• Choose Graph > Scatter plot.
• Choose With Connect Line, and then click OK.
• In Y variables, enter the Cumulative Percent.
• In X variables enter the Lower class limits.
• In Data view, select Symbols and Connect line under Data display.
• In Data view, select Smoother and enter 0 for Degree of smoothing and 1 for Number of steps under Lowness.
• Click OK
• To modify the interval settings, double click on the horizontal axis of the graph. Then, select Labels > Specified. In this box, enter the values for the cutpoints of the bin intervals (39.9, 42.9, 45.9, 48.9, 51.9, 54.9).

b.

To determine

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

Answer to Problem 3.3.23RE

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

Explanation of Solution

Calculation:

From part (a) the cumulative percent table is as follows,

Class limit	Frequency	Cumulative frequency	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
	${\displaystyle \sum f}=29$

For 35^th percentile:

The 35^th percentile location is calculated below,

Substitute n as 29 and m as 35 in the formula,

$\begin{array}{c}c=\frac{35\times 29}{100}\\ =\frac{1,015}{100}\\ =10.15\end{array}$

Here the 10^th observation corresponds to the cumulative frequency 14 which falls in the class 48.9-51.8.

The 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, the lower limit of the class.
• h, the width of class.
• f, the frequency of the class.
• p, the percentiles rank.
• n is the total number
• c is the preceding cumulative frequency

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.

For 65^th percentile:

The 65^th percentile location is calculated below,

Substitute n as 29 and m as 65 in the formula,

$\begin{array}{c}c=\frac{65\times 29}{100}\\ =\frac{1,885}{100}\\ =18.85\end{array}$

Here the 19^th observation corresponds to the cumulative frequency 26 which falls in the class 51.9-54.8.

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

For 85^th percentile:

The 85^th percentile location is calculated below,

Substitute n as 29 and m as 85 in the formula,

$\begin{array}{c}c=\frac{85\times 29}{100}\\ =\frac{2,465}{100}\\ =24.65\end{array}$

Here the 25^th observation corresponds to the cumulative frequency 26 which falls in the class 51.9-54.8.

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

c.

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 10^th, 26^th and 78^th respectively.

Explanation of Solution

Calculation:

Class limit	Frequency	Cumulative frequency
39.9-42.8	2	2
42.9-45.8	2	4
45.9-48.8	5	9
48.9-51.8	5	14
51.9-54.8	12	26
54.9-57.8	3	29
	${\displaystyle \sum f}=29$

For the value 44:

Here, the value 44 falls 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 formula,

      $\begin{array}{c}{P}_m=l+\frac{h}{f}\left(\frac{m.n}{100}-c\right)\\ 44=42.9+\frac{3}{2}\left(\frac{m\times 29}{100}-2\right)\\ 44=42.9+1.5\left(0.29m-2\right)\\ 44=42.9+0.435m-3\end{array}$

$\begin{array}{c}44=39.9+0.435m\\ 0.435m=44-39.9\\ m=\frac{4.1}{0.435}\\ =9.43\end{array}$

                 $\approx 10$

Thus, the percentile rank for the value 44 is 10^th percentile.

For the value 48:

Here, the value 48 falls in the interval 45.9-48.8

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,

       $\begin{array}{c}{P}_m=l+\frac{h}{f}\left(\frac{m.n}{100}-c\right)\\ 48=48.9+\frac{3}{5}\left(\frac{m\times 29}{100}-9\right)\\ 48=48.9+0.6\left(0.29m-9\right)\\ 48=48.9+0.174m-5.4\end{array}$

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

               $\approx 26$

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

For the value 48:

Here, the value 48 falls 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,

         $\begin{array}{c}{P}_m=l+\frac{h}{f}\left(\frac{m.n}{100}-c\right)\\ 54=51.9+\frac{3}{12}\left(\frac{m\times 29}{100}-14\right)\\ 54=51.9+\frac{3}{12}\left(0.29m-14\right)\\ 54=51.9+0.0725m-3.5\end{array}$

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

                 $\approx 78$

Thus, the percentile rank for the value 54 is 78^th percentile