Chapter 3.3, Problem 38E

a)

To determine

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

Answer to Problem 38E

First quartile = 1414.75 and third quartile = 2931

Explanation of Solution

Formula:

First quartile Q1:

  $Q_1=\frac{1}{4}{(n+1)}^{th}value$

Third quartile Q3:

  $Q_3=\frac{3}{4}{(n+1)}^{th}value$

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}value$$ Q_1=\frac{1}{4}{(59)}^{th}value$$ Q_1={(14.75)}^{th}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}value$$ Q_3=\frac{3}{4}{(58+1)}^{th}value$$ Q_3=\frac{3}{4}{(59)}^{th}value$$ Q_3={(44.25)}^{th}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 38E

Median is 2083.5

Explanation of Solution

Formula:

  $median={\left(\frac{n+1}{2}\right)}^{th}value$

Calculation:

  $\begin{array}{l}median={\left(\frac{58+1}{2}\right)}^{th}value\\ median={\left(29.5\right)}^{th}value\\ Median=\frac{2071+2096}{2}\\ Median=2083.5\end{array}$

c)

To determine

To find upper and lower outlier boundaries.

Answer to Problem 38E

Lower outlier boundary is -859.625

Upper outlier boundary is 5205.375

Explanation of Solution

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

Formula:

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,

  $\begin{array}{l}Loweroutlierlimit=Q_1-1.5\times IQR\hfill \\ Loweroutlierlimit=1414\text{-1}\text{.5}\times 1516.25\hfill \\ Loweroutlierlimit=\text{ -859}\text{.625}\hfill \\ And\hfill \\ Upperoutlierlimit=Q3+1.5\times IQR\hfill \\ Upperoutlierlimit=\text{ 2931+1}\text{.5}\times 1516.25\hfill \\ Upperoutlierlimit=\text{ 5205}\text{.375}\hfill \end{array}$

d)

To determine

To find the 135 and 559 are outliers or not.

Answer to Problem 38E

135 and 559 are not outliers.

Explanation of Solution

Outliers are those values which are less than Q1-1.5 x IQR and greater than Q3+1.5 x 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 38E

The 8460 and 5434 are outliers.

Explanation of Solution

Outliers are those values which are less than Q1-1.5 x IQR and greater than Q3+1.5 x 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 38E

The shape of the distribution is positively-skewed.

Explanation of Solution

Here, median is close to first quartile. Hence the shape of the distribution is positively-skewed.

h)

To determine

To find 15^st percentile

Answer to Problem 38E

15^st percentile is 1176.85

Explanation of Solution

  $\begin{array}{l}PercentilePosition=\frac{(n+1)P}{100}\\ PercentilePosition=\frac{(58+1)\times 15}{100}\\ PercentilePosition=8.85\end{array}$

Therefore,

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

i)

To determine

To find 65^th percentile

Answer to Problem 38E

65^th percentile is 2406.2

Explanation of Solution

  $\begin{array}{l}PercentilePosition=\frac{(n+1)P}{100}\\ PercentilePosition=\frac{(58+1)\times 65}{100}\\ PercentilePosition=38.35\end{array}$

Therefore,

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

j)

To determine

To find percentile rank for 1433 words

Answer to Problem 38E

Percentile rank for 1433 words is 25.86%

Explanation of Solution

Formula:

  $Percentilerankofx=\frac{no.ofvaluesbelowx}{n}\times 100$

Calculation:

  $\begin{array}{l}Percentilerankof1433=\frac{15}{58}\times 100\\ Percentilerankof1433=25.86\%\end{array}$