Chapter 3.3, Problem 18E

College Room and Board Costs Room and board costs for selected schools are summarized in this distribution. Find the approximate cost of room and board corresponding to each of the following percentiles.

Costs (in dollars)	Frequency
3000.5–4000.5 4000.5–5000.5 5000.5–6000.5 6000.5–7000.5 7000.5–8000.5 8000.5–9000.5 9000.5–10,000.5	5 6 18 24 19 8 5

a. 30th percentile

b. 50th percentile

c. 75th percentile

d. 90th percentile

Source: World Almanac.

Using the same data, find the approximate percentile rank of each of the following costs.

e. 5500

f. 7200

g. 6500

h. 8300

To determine

The 30^th percentile of the given data.

Answer to Problem 18E

The 30^th percentile of the given data is $5,06.05.

Explanation of Solution

Given info:

Given data is tabulated below:

Costs (in dollars)	Frequency
3,000.5-4,000.5	5
4,000.5-5,000.5	6
5,000.5-6,000.5	18
6,000.5-7,000.5	24
7,000.5-8,000.5	19
8,000.5-9,000.5	8
9,000.5-10,000.5	5

Calculation:

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

A Class

B Frequency( $f$)

C Cumulative frequency( $cf$)

D Cumulative $cf\%$

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

Determine percentage of cumulative frequency and write in column D.

Table for calculating percentiles given below,

A Class	B Frequency( $f$)	C Cumulative frequency( $cf$)	D Cumulative $cf\%$
3,000.5-4,000.5	5	5	5.88
4,000.5-5,000.5	6	11	12.94
5,000.5-6,000.5	18	29	34.12
6,000.5-7,000.5	24	53	62.35
7,000.5-8,000.5	19	72	84.71
8,000.5-9,000.5	8	80	94.12
9,000.5-10,000.5	5	85	100
Total	85

Locate the $30\text{th}$ percentile by $\frac{30\times 85}{100}=25.5$ observation. So the $30\text{th}$ percentiles group is 5,000.5-6,000.5 containing the $25.5\text{th}$ observation.

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.
• $n$ is the total number.
• $c$ is preceding cumulative frequency.

Substitute 5,000.5 for the $l$ and 1,000 for the $h$ and 18 for $f$ and 30 for $p$ and 85 for $n$ and 11 for $c$,

$\begin{array}{c}{P}_{30}=5,000.5+\frac{1,000}{18}\left(\frac{30\times 85}{100}-11\right)\\ =5,000.5+\frac{1,000}{18}\left(25.5-11\right)\\ =5,000.5+\frac{1,000}{18}\left(14.5\right)\\ =5,000.5+\frac{1,000\times 14.5}{18}\end{array}$

Simplify the equation,

$\begin{array}{c}{P}_{30}=5,000.5+805.56\\ =5,806.05\end{array}$

Thus, the 30^th percentile of the data is 5,806.05.

To determine

The $\text{50th}$ percentiles of the given data.

Answer to Problem 18E

The $\text{50th}$ percentiles of the given data is6563.

Explanation of Solution

Calculation:

Substitute 6,000.5 for the $l$ and 1,000 for the $h$ and 24 for $f$ and 50 for $p$ and 85 for $n$ and 29 for $c$,

$\begin{array}{c}{P}_{50}=6,000.5+\frac{1,000}{24}\left(\frac{50\times 85}{100}-29\right)\\ =6,000.5+\frac{1,000}{24}\left(42.5-29\right)\\ =6,000.5+\frac{1,000}{24}\left(13.5\right)\\ =6,000.5+\frac{1,000\times 13.5}{24}\end{array}$

Simplify the equation.

$\begin{array}{c}{P}_{50}=6,000.5+562.5\\ =6,563\end{array}$

Thus the 50^th percentile of the data is 6,563.

To determine

The 75^th percentile of the given data.

Answer to Problem 18E

The 75^th percentile of the given data is 7,566.29.

Explanation of Solution

Calculation:

Substitute 7,000.5 for the $l$ and 1,000 for the $h$ and 19 for $f$ and 75 for $p$ and 85 for $n$ and 53 for $c$.

$\begin{array}{c}{P}_{75}=7,000.5+\frac{1,000}{19}\left(\frac{75\times 85}{100}-53\right)\\ =7,000.5+\frac{1,000}{19}\left(63.75-53\right)\\ =7,000.5+\frac{1,000}{19}\left(10.75\right)\\ =7,000.5+\frac{1,000\times 10.75}{19}\end{array}$

Simplify the equation,

$\begin{array}{c}{P}_{75}=7,000.5+565.79\\ =7,566.29\end{array}$

Thus, the 75^th percentile of the data is 7,566.29.

To determine

The 90^th percentile of the given data.

Answer to Problem 18E

The 90^th percentile of the given data is 8,563.

Explanation of Solution

Calculation:

Substitute 8,000.5 for the $l$ and 1,000 for the $h$ and 8 for $f$ and 90 for $p$ and 85 for $n$ and 72 for $c$.

$\begin{array}{c}{P}_{90}=8,000.5+\frac{1,000}{8}\left(\frac{90\times 85}{100}-72\right)\\ =8,000.5+\frac{1,000}{8}\left(76.5-72\right)\\ =8,000.5+\frac{1,000}{8}\left(4.5\right)\\ =8,000.5+\frac{1,000\times 4.5}{8}\end{array}$

Simplify the equation,

$\begin{array}{c}{P}_{90}=8,000.5+562.5\\ =8,563\end{array}$

Thus the 90^th percentile of the data is 8,563.

To determine

The percentiles rank corresponding to value 5500.

Answer to Problem 18E

The 24^th percentile rank corresponding to the value is 5,500.

Explanation of Solution

Calculation:

Substitute 5,000.5 for the $l$ and 10,00 for the $h$ and 18 for $f$ and 5500 for $P$ and 85 for $n$ and 11 for $c$.

$\begin{array}{c}5,500=5,000.5+\frac{1,000}{18}\left(\frac{p\times 85}{100}-11\right)\\ =5000.5+\frac{1,000}{18}\left(0.85p-11\right)\\ =5,000.5+55.56\left(0.85p-11\right)\\ =5,000.5+47.23p-611.16\end{array}$

Simplify the equation,

$\begin{array}{c}47.23p=5500-4389.34\\ p=\frac{1110.66}{47.23}\\ =23.5\\ \approx 24\end{array}$

Thus the percentiles rank for the cost of 5500 is $24\text{th}$ percentiles.

To determine

The percentiles rank corresponding to value 7200.

Answer to Problem 18E

The $\text{67th}$ percentiles rank corresponding to the value 7200.

Explanation of Solution

Calculation:

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

A Class

B Frequency( $f$)

C Cumulative frequency( $cf$)

D Cumulative $cf\%$

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

Determine percentage of cumulative frequency and write in column D.

Table for calculating percentiles given below,

A Class	B Frequency( $f$)	C Cumulative frequency( $cf$)	D Cumulative $cf\%$
3000.5-4000.5	5	5	5.88
4000.5-5000.5	6	11	12.94
5000.5-6000.5	18	29	34.12
6000.5-7000.5	24	53	62.35
7000.5-8000.5	19	72	84.71
8000.5-9000.5	8	80	94.12
9000.5-10000.5	5	85	100
${\displaystyle \sum }$	85

The cost 7200 lie in the interval 7000.5-8000.5

Substitute 7000.5 for the $l$ and 1000 for the $h$ and 19 for $f$ and 7200 for $P$ and 85 for $n$ and 53 for $c$.

$\begin{array}{c}7200=7000.5+\frac{1000}{19}\left(\frac{p\times 85}{100}-53\right)\\ =7000.5+\frac{1000}{19}\left(0.85p-53\right)\\ =7000.5+52.63\left(0.85p-53\right)\\ =7000.5+44.74p-2789.39\end{array}$

Simplify the equation,

$\begin{array}{c}44.74p=7200-4211.11\\ p=\frac{2988.89}{44.74}\\ =66.8\\ \approx 67\left(\text{rounding}\text{up}\right)\end{array}$

Thus the percentiles rank for the cost of 7200 is $\text{67th}$ percentiles.

To determine

The percentiles rank corresponding to value 6500.

Answer to Problem 18E

The $\text{48th}$ percentiles rank corresponding to the value 6500.

Explanation of Solution

Calculation:

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

A Class

B Frequency( $f$)

C Cumulative frequency( $cf$)

D Cumulative $cf\%$

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

Determine percentage of cumulative frequency and write in column D.

Table for calculating percentiles given below,

A Class	B Frequency( $f$)	C Cumulative frequency( $cf$)	D Cumulative $cf\%$
3000.5-4000.5	5	5	5.88
4000.5-5000.5	6	11	12.94
5000.5-6000.5	18	29	34.12
6000.5-7000.5	24	53	62.35
7000.5-8000.5	19	72	84.71
8000.5-9000.5	8	80	94.12
9000.5-10000.5	5	85	100
Total	85

The cost 6500 lie in the interval 6000.5-7000.5.

Substitute 6000.5 for the $l$ and 1000 for the $h$ and 24 for $f$ and 6500 for $P$ and 85 for $n$ and 29 for $c$,

$\begin{array}{c}6500=6000.5+\frac{1000}{24}\left(\frac{p\times 85}{100}-29\right)\\ =6000.5+\frac{1000}{24}\left(0.85p-29\right)\\ =6000.5+41.67\left(0.85p-29\right)\\ =8000.5+435.42p-1208.43\end{array}$

Simplify the equation,

$\begin{array}{c}35.42p=6500-4792.07\\ p=\frac{1707.93}{35.42}\\ =48.22\\ \approx 48\text{th}\end{array}$

Thus the percentiles rank for the cost of 6500 is $\text{48th}$ percentiles.

To determine

The percentiles rank corresponding to value 8300.

Answer to Problem 18E

The $\text{88th}$ percentiles rank corresponding to the value 8300.

Explanation of Solution

Calculation:

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

A Class

B Frequency( $f$)

C Cumulative frequency( $cf$)

D Cumulative $cf\%$

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

Determine percentage of cumulative frequency and write in column D.

Table for calculating percentiles given below,

A Class	B Frequency( $f$)	C Cumulative frequency( $cf$)	D Cumulative $cf\%$
3000.5-4000.5	5	5	5.88
4000.5-5000.5	6	11	12.94
5000.5-6000.5	18	29	34.12
6000.5-7000.5	24	53	62.35
7000.5-8000.5	19	72	84.71
8000.5-9000.5	8	80	94.12
9000.5-10000.5	5	85	100
${\displaystyle \sum }$	85

The cost 8300 lies in the interval 8000.5-9000.5

Substitute 8000.5 for the $l$ and 1000 for the $h$ and 8 for $f$ and 8300 for $P$ and 85 for $n$ and 72 for $c$.

$\begin{array}{c}8300=8000.5+\frac{1000}{8}\left(\frac{p\times 85}{100}-72\right)\\ =8000.5+\frac{1000}{8}\left(0.85p-72\right)\\ =8000.5+125\left(0.85p-72\right)\\ =8000.5+106.25p-9000\end{array}$

Simplify the equation,

$\begin{array}{c}106.25p=8300-8000.5+9000\\ p=\frac{9299.5}{106.25}\\ =87.5\\ \approx 88\end{array}$

Thus the percentiles rank for the cost of 8300 is 88th percentile.