Chapter 3.3, Problem 19E

a.

To determine

The percentile rank corresponding to value 220.

Answer to Problem 19E

The percentile rank corresponding to the value 220 is 6^th percentile.

Explanation of Solution

Given info:

The data represents the scores on a national achievement test for a group of 10ht-grade students.

Score	Frequency
196.5-217.5	5
217.5-238.5	17
238.5-259.5	22
259.5-280.5	48
280.5-301.5	22
301.5-322.5	6

Calculation:

The cumulative frequencies of the distribution are calculated below,

Score	Frequency f	Cumulative frequency
196.5-217.5	5	5
217.5-238.5	17	22
238.5-259.5	22	44
259.5-280.5	48	92
280.5-301.5	22	114
301.5-322.5	6	120
Total	${\displaystyle \sum f}=120$

Here, the value 220falls in the interval 217.5-238.5.

The formula to calculate percentile for the correspondingvalue is as follows,

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

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 l as 217.5, h as 21, f as 17, P as 220, n as 120, and c as 5 in the formula,

$\begin{array}{c}P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)\\ 220=217.5+\frac{21}{17}\left(\frac{p\times 120}{100}-5\right)\\ 220=217.5+\frac{21}{17}\left(1.2p-5\right)\\ 220=217.5+1.2353\left(1.2p-5\right)\end{array}$

$\begin{array}{c}220=217.5+1.4824p-6.1765\\ 220=211.3235+1.4824p\\ 1.4824p=220-211.3235\\ p=\frac{8.6765}{1.4824}\end{array}$

$\begin{array}{c}=5.8530\\ \approx 6\left(\text{rounding}\text{up}\right)\end{array}$

Therefore, the percentile rank for 220 is 6^th percentile.

b.

To determine

The percentile rank corresponding to value 245.

Answer to Problem 19E

The percentile rank that corresponds to 245 is 24^th percentile.

Explanation of Solution

Calculation:

Here, the value 245falls in the interval 238.5-259.5.

The formula to calculate percentile for the correspondingvalue is as follows,

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

Substitute l as 238.5, h as 21, f as 22, P as 245, n as 120, and c as 22 in the formula,

$\begin{array}{c}P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)\\ 245=238.5+\frac{21}{22}\left(\frac{p\times 120}{100}-22\right)\\ 245=238.5+\frac{21}{22}\left(1.2p-22\right)\\ 245=238.5+0.9546\left(1.2p-22\right)\end{array}$

$\begin{array}{c}245=238.5+1.1455p-21.0012\\ 245=217.4988+1.1455p\\ 1.1455p=245-217.4988\\ p=\frac{27.5012}{1.1455}\end{array}$

$\begin{array}{c}=24.0080\\ \approx 24\end{array}$

Therefore, the percentile rank that corresponds to 245 is 24^th percentile.

c.

To determine

The percentile rank corresponding to value 276.

Answer to Problem 19E

The percentile rank that corresponds to 276 is 68^th percentile.

Explanation of Solution

Calculation:

Here, the value 276falls in the interval 259.5-280.5.

The formula to calculate percentile for the correspondingvalue is as follows,

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

Substitute l as 259.5, h as 21, f as 48, P as 276, n as 120, and c as 44 in the formula,

$\begin{array}{c}P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)\\ 276=259.5+\frac{21}{48}\left(\frac{p\times 120}{100}-44\right)\\ 276=259.5+\frac{21}{48}\left(1.2p-44\right)\\ 276=259.5+0.4375\left(1.2p-44\right)\end{array}$

$\begin{array}{c}276=259.5+0.525p-19.25\\ 276=240.25+0.525p\\ 0.525p=276-240.25\\ p=\frac{35.75}{0.525}\end{array}$

$\begin{array}{c}=68.0952\\ \approx 68\end{array}$

Therefore, the percentile rank that corresponds to 276 is 68^th percentile.

d.

To determine

The percentile rank corresponding to value 280.

Answer to Problem 19E

The percentile rank that corresponds to 280 is 76^th percentile.

Explanation of Solution

Calculation:

Here, the value 280falls in the interval 280.5-301.5.

The formula to calculate percentile for the correspondingvalue is as follows,

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

Substitute l as 280.5, h as 21, f as 22, P as 280, n as 120, and c as 92 in the formula,

$\begin{array}{c}P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)\\ 280=280.5+\frac{21}{22}\left(\frac{p\times 120}{100}-92\right)\\ 280=280.5+\frac{21}{22}\left(1.2p-92\right)\\ 280=280.5+0.9546\left(1.2p-92\right)\end{array}$

$\begin{array}{c}280=280.5+1.1455p-87.8232\\ 280=192.6768+1.1455p\\ 1.1455p=280-192.6768\\ p=\frac{87.3232}{1.1455}\end{array}$

$\begin{array}{c}=76.2315\\ \approx 76\end{array}$

Therefore, the percentile rank that corresponds to 280 is 76^th percentile.

e.

To determine

The percentile rank corresponding to value 300.

Answer to Problem 19E

The percentile rank that corresponds to 300 is 94^th percentile.

Explanation of Solution

Calculation:

Here, the value 300falls in the interval 280.5-301.5.

The formula to calculate percentile for the correspondingvalue is as follows,

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

Substitute l as 280.5, h as 21, f as 22, P as 280, n as 120, and c as 92 in the formula,

$\begin{array}{c}P=l+\frac{h}{f}\left(\frac{p\times n}{100}-c\right)\\ 300=280.5+\frac{21}{22}\left(\frac{p\times 120}{100}-92\right)\\ 300=280.5+\frac{21}{22}\left(1.2p-92\right)\\ 300=280.5+0.9546\left(1.2p-92\right)\end{array}$

$\begin{array}{c}300=280.5+1.1455p-87.8232\\ 300=192.6768+1.1455p\\ 1.1455p=300-192.6768\\ p=\frac{107.3232}{1.1455}\end{array}$

$\begin{array}{c}=93.6911\\ \approx 94\end{array}$

Therefore, the percentile rank that corresponds to 300 is 94^th percentile.

f.

To determine

To find: The 15^th percentile of the data.

Answer to Problem 19E

The 15^th percentile of the data is 234.

Explanation of Solution

Calculation:

The cumulative frequencies of the distribution are calculated below,

Score	Frequency f	Cumulative frequency
196.5-217.5	5	5
217.5-238.5	17	22
238.5-259.5	22	44
259.5-280.5	48	92
280.5-301.5	22	114
301.5-322.5	6	120
Total	${\displaystyle \sum f}=120$

The formula to locate percentile is as follows,

$c=\frac{n\cdot p}{100}$

The 15^th percentile location:

Substitute n as 120 and p as 15 in the formula,

$\begin{array}{c}c=\frac{120\times 15}{100}\\ =\frac{1800}{100}\\ =18\end{array}$

Here the 18^th observation corresponds to the cumulative frequency 22 which falls in the class 217.5-238.5.

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 l as 217.5, h as 21,f as 17, p as 15, n as 120 and c as 5.

$\begin{array}{c}{P}_{15}=217.5+\frac{21}{17}\left(\frac{120\times 15}{100}-5\right)\\ =217.5+\frac{21}{17}\left(18-5\right)\\ =217.5+\frac{21}{17}\left(13\right)\\ =217.5+16.0588\end{array}$

$\begin{array}{c}=233.5588\\ \approx 234\end{array}$

Therefore, the 15^th percentile of the data is 234.

g.

To determine

The 29^th percentile of the given data.

Answer to Problem 19E

The 29^th percentile of the given data is 251.

Explanation of Solution

Calculation:

The cumulative frequencies of the distribution are calculated below,

Score	Frequency f	Cumulative frequency
196.5-217.5	5	5
217.5-238.5	17	22
238.5-259.5	22	44
259.5-280.5	48	92
280.5-301.5	22	114
301.5-322.5	6	120
Total	${\displaystyle \sum f}=120$

The formula to locate percentile is as follows,

$c=\frac{n\cdot p}{100}$

The 29^th percentile location:

Substitute n as 120 and p as 29 in the formula,

$\begin{array}{c}c=\frac{120\times 29}{100}\\ =\frac{3,480}{100}\\ =34.8\\ \approx 35\end{array}$

Here the 35^th observation corresponds to the cumulative frequency 44 which falls in the class 238.5-259.5.

Formula to calculate value corresponding percentiles is,

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

Substitute l as 238.5, h as 21,f as 22, p as 29, n as 120 and c as 22.

$\begin{array}{c}{P}_{29}=238.5+\frac{21}{22}\left(\frac{120\times 29}{100}-22\right)\\ =238.5+\frac{21}{22}\left(34.8-22\right)\\ =238.5+\frac{21}{22}\left(12.8\right)\\ =238.5+12.2182\end{array}$

$\begin{array}{c}=250.7182\\ \approx 251\end{array}$

Therefore, the 29^th percentile of the data is 251.

h.

To determine

The 43^rd percentile of the given data.

Answer to Problem 19E

The 43^rd percentile of the given data is 263.

Explanation of Solution

Calculation:

The cumulative frequencies of the distribution are calculated below,

Score	Frequency f	Cumulative frequency
196.5-217.5	5	5
217.5-238.5	17	22
238.5-259.5	22	44
259.5-280.5	48	92
280.5-301.5	22	114
301.5-322.5	6	120
Total	${\displaystyle \sum f}=120$

The formula to locate percentile is as follows,

$c=\frac{n\cdot p}{100}$

The 43^rd percentile location:

Substitute n as 120 and p as 43 in the formula,

$\begin{array}{c}c=\frac{120\times 43}{100}\\ =\frac{5,160}{100}\\ =51.6\\ \approx 52\end{array}$

Here the 52^nd observation corresponds to the cumulative frequency 92 which falls in the class 259.5-280.5.

Formula to calculate value corresponding percentiles is,

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

Substitute l as 259.5, h as 21,f as 48, p as 43, n as 120 and c as 44.

$\begin{array}{c}{P}_{43}=259.5+\frac{21}{48}\left(\frac{120\times 43}{100}-44\right)\\ =259.5+\frac{21}{48}\left(51.6-44\right)\\ =259.5+\frac{21}{48}\left(7.6\right)\\ =259.5+3.325\end{array}$

$\begin{array}{c}=262.825\\ \approx 263\end{array}$

Therefore, the 43^rd percentile of the data is 263.

i.

To determine

The 65^th percentile of the given data.

Answer to Problem 19E

The 65^th percentile of the given data is 274.

Explanation of Solution

Calculation:

The cumulative frequencies of the distribution are calculated below,

Score	Frequency f	Cumulative frequency
196.5-217.5	5	5
217.5-238.5	17	22
238.5-259.5	22	44
259.5-280.5	48	92
280.5-301.5	22	114
301.5-322.5	6	120
Total	${\displaystyle \sum f}=120$

The formula to locate percentile is as follows,

$c=\frac{n\cdot p}{100}$

The 65^th percentile location:

Substitute n as 120 and p as 65 in the formula,

$\begin{array}{c}c=\frac{120\times 65}{100}\\ =\frac{7,800}{100}\\ =78\end{array}$

Here the 78^th observation corresponds to the cumulative frequency 92 which falls in the class 259.5-280.5.

Formula to calculate value corresponding percentiles is,

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

Substitute l as 259.5, h as 21,f as 48, p as 65, n as 120 and c as 44.

$\begin{array}{c}{P}_{65}=259.5+\frac{21}{48}\left(\frac{120\times 65}{100}-44\right)\\ =259.5+\frac{21}{48}\left(78-44\right)\\ =259.5+\frac{21}{48}\left(34\right)\\ =259.5+14.875\end{array}$

$\begin{array}{c}=274.375\\ \approx 274\end{array}$

Therefore, the 65^th percentile of the data is 274.

j.

To determine

The 80^th percentile of the given data.

Answer to Problem 19E

The 80^th percentile of the given data is 284.

Explanation of Solution

Calculation:

The cumulative frequencies of the distribution are calculated below,

Score	Frequency f	Cumulative frequency
196.5-217.5	5	5
217.5-238.5	17	22
238.5-259.5	22	44
259.5-280.5	48	92
280.5-301.5	22	114
301.5-322.5	6	120
Total	${\displaystyle \sum f}=120$

The formula to locate percentile is as follows,

$c=\frac{n\cdot p}{100}$

The 80^th percentile location:

Substitute n as 120 and p as 80 in the formula,

$\begin{array}{c}c=\frac{120\times 80}{100}\\ =\frac{9,600}{100}\\ =96\end{array}$

Here the 96^th observation corresponds to the cumulative frequency 114 which falls in the class 280.5-301.5.

Formula to calculate value corresponding percentiles is,

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

Substitute l as 280.5, h as 21,f as 22, p as 80, n as 120 and c as 92.

$\begin{array}{c}{P}_{80}=280.5+\frac{21}{22}\left(\frac{120\times 80}{100}-92\right)\\ =280.5+\frac{21}{22}\left(96-92\right)\\ =280.5+\frac{21}{22}\left(4\right)\\ =280.5+3.8182\end{array}$

$\begin{array}{c}=284.3182\\ \approx 284\end{array}$

Therefore, the 80^th percentile of the data is 284.