Chapter 3.3, Problem 32E

(a)

To determine

To find the first and third quartiles of the given data.

Answer to Problem 32E

First quartile = 1414.75 and third quartile = 2931

Explanation of Solution

Formula used:

First quartile $Q_1=\frac{1}{4}{(n+1)}^{th}value$

If $\frac{1}{4}(n+1)$ is not an integer, then

$Q_1=x_{\mathrm{int}eger\left(\frac{1}{4}(n+1)\right)}+\left(x_{\mathrm{int}eger\left(\frac{1}{4}(n+1)\right)+1}-x_{\mathrm{int}eger\left(\frac{1}{4}(n+1)\right)}\right)\left(decimal\left(\left(\frac{1}{4}(n+1)\right)\right)\right)$

Third quartile $Q_3=\frac{3}{4}{(n+1)}^{th}value$

If $\frac{3}{4}(n+1)$ is not an integer, then

$Q_1=x_{\mathrm{int}eger\left(\frac{3}{4}(n+1)\right)}+\left(x_{\mathrm{int}eger\left(\frac{3}{4}(n+1)\right)+1}-x_{\mathrm{int}eger\left(\frac{3}{4}(n+1)\right)}\right)\left(decimal\left(\left(\frac{3}{4}(n+1)\right)\right)\right)$

Calculation:

Data sorted in ascending order:

$x$
135	1229	1658	2128	2486	3843
559	1339	1686	2155	2561	3968
700	1359	1704	2166	2831	4055
984	1366	1730	2218	2915	4392
1090	1431	1803	2273	2979	4472
1127	1433	1808	2320	3329	4809
1128	1507	1880	2321	3336	5434
1176	1526	2015	2395	3375	8460
1177	1592	2071	2427	3637
1211	1598	2096	2459	3672

Here, n = 58

First need to find First quartile and third quartile

First Quartile:

$\begin{array}{l}$ Q_1=\frac{1}{4}{(58+1)}^{th}\text{value}$$ Q_1=\frac{1}{4}{(59)}^{th}\text{value}$$ Q_1={(14.75)}^{th}\text{value}$Q_1=1366+0.75\times (1431-1366)\\ Q_1=1414.75\end{array}$

First quartile is 1414.75

Third quartile:

$\begin{array}{l}$ Q_3=\frac{3}{4}{(n+1)}^{th}\text{value}$$ Q_3=\frac{3}{4}{(58+1)}^{th}\text{value}$$ Q_3=\frac{3}{4}{(59)}^{th}\text{value}$$ Q_3={(44.25)}^{th}\text{value}$Q_3=2915+0.25\times (2979-2915)\\ Q_3=2931\end{array}$

Third quartile is 2931.

(b)

To determine

To find median of the data.

Answer to Problem 32E

Median is 2083.5

Explanation of Solution

Formula used:

$\begin{array}{c}\text{Median}={\left(\frac{n+1}{2}\right)}^{th}\text{value}\text{for}\text{odd}n\\ =\frac{1}{2}\left[{\left(\frac{n}{2}\right)}^{th}\text{value}+{\left(\frac{n}{2}+1\right)}^{th}\text{value}\right]\text{for}\text{even}n\end{array}$

Calculation:

Here $n=58$, which is even

$\begin{array}{c}\text{Median}=\frac{1}{2}\left[{\left(\frac{58}{2}\right)}^{th}\text{value}+{\left(\frac{58}{2}+1\right)}^{th}\text{value}\right]\\ =\frac{1}{2}\left[{\left(29\right)}^{th}\text{value}+{\left(30\right)}^{th}\text{value}\right]\\ =\frac{2071+2096}{2}\\ =2083.5\end{array}$

(c)

To determine

To find upper and lower outlier boundaries.

Answer to Problem 32E

Lower outlier boundary is -859.625

Upper outlier boundary is 5205.375

Explanation of Solution

Given:

From part (a)

$\begin{array}{l}{Q}_1=1414.75\\ Q_3=2931\end{array}$

Formula used:

IQR:

$IQR=Q_3-Q_1$

Calculation:

$\begin{array}{l}IQR=Q_3-Q_1\hfill \\ IQR=\text{ 2931-1414}\text{.75}\hfill \\ IQR=\text{ 1516}\text{.25}\hfill \end{array}$

Therefore,

$\begin{array}{c}\text{Lower outlier limit }=Q_1-1.5\times IQR\\ =1414\text{-1}\text{.5}\times 1516.25\\ =-\text{859}\text{.625}\end{array}$

And

$\begin{array}{c}\text{Upper outlier limit }=Q3+1.5\times IQR\\ =\text{2931+1}\text{.5}\times 1516.25\\ \text{= 5205}\text{.375}\end{array}$

d)

To determine

To find the 135 and 559 are outliers or not.

Answer to Problem 32E

135 and 559 are not outliers.

Explanation of Solution

Outliers are those values which are less than $Q_1-1.5\times IQR$ and greater than $Q_3+1.5\times IQR$.

Here,

Lower outlier boundary is -859.625

Upper outlier boundary is 5205.375

135 and 559 are within range of Lower and upper outlier boundary.

(e)

To determine

To find the 8460 and 5434 are outliers or not.

Answer to Problem 32E

The 8460 and 5434 are outliers.

Explanation of Solution

Outliers are those values which are less than $Q_1-1.5\times IQR$ and greater than $Q_3+1.5\times IQR$.

Here,

Lower outlier boundary is -859.625

Upper outlier boundary is 5205.375

8460 and 5434 are greater than upper outlier limit 5205.375. Hence both are outliers.

(f)

To determine

To construct a boxplot from given data.

Explanation of Solution

Boxplot from given data:

(g)

To determine

To find shape of the distribution.

Answer to Problem 32E

The shape of the distribution is Positively skewed.

Explanation of Solution

Since the difference between the median and third quartile is larger than the difference between first quartile and median. Which also means that median is close to first quartile therefore the shape of distribution is right skewed or positively skewed.

(h)

To determine

To find 15^{s t} percentile.

Answer to Problem 32E

15^{s t} percentile is 1176.85

Explanation of Solution

Calculation:

$\begin{array}{c}\text{Percentile}\text{Position}=\frac{(n+1)P}{100}\\ =\frac{(58+1)\times 15}{100}\\ =8.85\end{array}$

Therefore, the percentile can be calculated as

$\begin{array}{c}{P}_{15}=1176+0.85\times (1177-1176)\\ =1176.85\end{array}$

(i)

To determine

To find 65^{t h} percentile.

Answer to Problem 32E

65^{t h} percentile is 2406.2

Explanation of Solution

Calculation:

$\begin{array}{c}\text{Percentile}\text{Position}=\frac{(n+1)P}{100}\\ =\frac{(58+1)\times 65}{100}\\ =38.35\end{array}$

Therefore,

$\begin{array}{c}{P}_{65}=2395+0.35\times (2427-2395)\\ =2406.2\end{array}$

(j)

To determine

To find percentile rank for 1433 words.

Answer to Problem 32E

Percentile rank for 1433 words is 25.86%.

Explanation of Solution

Formula used:

$\text{Percentile}\text{rank}\text{of}x=\frac{\text{no}\text{.of}\text{values}\text{below}x}{n}\times 100$

Calculation:

Number of values below 1433 is 15.

$\begin{array}{c}\text{Percentile}\text{rank}\text{of}1433=\frac{15}{58}\times 100\\ =25.86\%\end{array}$