Chapter 14.3, Problem 20E

a.

To determine

To calculate: The median of the lengths.

Answer to Problem 20E

The median is $259$ .

Explanation of Solution

Given information:

There are seven navigable rivers that feed into the Ohio river. The lengths of these rivers are

  

Formula used:

Median for even terms = $\frac{{\left(\frac{n}{2}\right)}^{th}\text{term}+{\left(\frac{n}{2}+1\right)}^{th}\text{term}}{2}$

Median for odd terms = ${\left(\frac{n+1}{2}\right)}^{th}\text{term}$

Calculation:

Consider the data in ascending order ,

  $\left\{97,129,169,259,325,360,694\right\}.$

Since there are seven rivers. So, $n=7$ .

There is odd number of data.

The median of odd number of data is −

  $\begin{array}{l}={\left(\frac{7+1}{2}\right)}^{th}\text{term}\\ \text{=}{\left(\frac{8}{2}\right)}^{th}\text{term}\\ =4^{th}\text{term}\\ =259\end{array}$

Hence, the median is $259$ .

b.

To determine

To calculate: The first quartile point and the third quartile point.

Answer to Problem 20E

The first quartile and third quartile are $129\text{ and }360\text{ respectively}\text{.}$

Explanation of Solution

Given information:

There are seven navigable rivers that feed into the Ohio river. The lengths of these rivers are

  

Formula used:

If each group has the median, the data is divided into four groups. Each group is called quartile.

Median for even terms = $\frac{{\left(\frac{n}{2}\right)}^{th}\text{term}+{\left(\frac{n}{2}+1\right)}^{th}\text{term}}{2}$

Median for odd terms = ${\left(\frac{n+1}{2}\right)}^{th}\text{term}$

Calculation:

Consider the data in ascending order ,

  $\left\{97,129,169,259,325,360,694\right\}.$

Since there are seven rivers. So, $n=7$ .

There is odd number of data.

The median of odd number of data is −

  $\begin{array}{l}={\left(\frac{7+1}{2}\right)}^{th}\text{term}\\ \text{=}{\left(\frac{8}{2}\right)}^{th}\text{term}\\ =4^{th}\text{term}\\ =259\end{array}$

Therefore , the data is divided into two parts.

Lower half - $97,129,169.$ and Upper half - $325,360,694.$

Median of lower half is −

  $\begin{array}{l}={\left(\frac{3+1}{2}\right)}^{th}\text{term}\\ ={\left(\frac{4}{2}\right)}^{th}\text{term}\\ ={\left(2\right)}^{th}\text{term}\\ =129\end{array}$

Median of upper half is −

  $\begin{array}{l}={\left(\frac{3+1}{2}\right)}^{th}\text{term}\\ ={\left(\frac{4}{2}\right)}^{th}\text{term}\\ ={\left(2\right)}^{th}\text{term}\\ =360\end{array}$

The quartile points are :-

  $\begin{array}{l}{Q}_1=129\\ Q_2=259\\ Q_3=360\end{array}$

Hence, the first quartile and third quartile are $129\text{ and }360\text{ respectively}\text{.}$

c.

To determine

To calculate: The interquartile range.

Answer to Problem 20E

The interquartile range is $231.$

Explanation of Solution

Given information:

There are seven navigable rivers that feed into the Ohio river. The lengths of these rivers are

  

Formula used:

The difference between the first quartile point and third quartile point is called the inter- quartile range.

Calculation:

Consider the data in ascending order ,

  $\left\{97,129,169,259,325,360,694\right\}.$

Since there are seven rivers. So, $n=7$ .

There is odd number of data.

The median of odd number of data is $259.$

The quartile points are :-

  $\begin{array}{l}{Q}_1=129\\ Q_2=259\\ Q_3=360\end{array}$

The inter- quartile range is $Q_3-Q_1.$

  $\begin{array}{l}=360-129\\ =231.\end{array}$

Hence, the interquartile range is $231.$

d.

To determine

To calculate: The semi-interquartile range.

Answer to Problem 20E

The semi-interquartile range is $\mathrm{115.5.}$

Explanation of Solution

Given information:

There are seven navigable rivers that feed into the Ohio river. The lengths of these rivers are

  

Formula used:

The inter- quartile range is divided by 2, the quotient is called the semi- interquartile range.

Calculation:

Consider the data in ascending order ,

  $\left\{97,129,169,259,325,360,694\right\}.$

Since there are seven rivers. So, $n=7$ .

There is odd number of data.

The median of odd number of data is $259.$

The quartile points are :-

  $\begin{array}{l}{Q}_1=129\\ Q_2=259\\ Q_3=360\end{array}$

The inter- quartile range is $231.$

The semi-interquartile range -

  $\frac{\text{inter-quartile range}}{2}.$

  $\begin{array}{l}=\frac{231}{2}\\ =\mathrm{115.5.}\end{array}$

Hence, the semi-interquartile range is $\mathrm{115.5.}$

e.

To determine

To show: whether there is any outliers.

Answer to Problem 20E

There are no outliers.

Explanation of Solution

Given information:

There are seven navigable rivers that feed into the Ohio river. The lengths of these rivers are

  

Formula used:

Outliers are extreme values that are more than 1.5 times the interquartile range beyond the upper or lower quartiles .

Calculation:

Consider the data in ascending order ,

  $\left\{97,129,169,259,325,360,694\right\}.$

Since there are seven rivers. So, $n=7$ .

There is odd number of data.

The median of odd number of data is $259.$

The quartile points are :-

  $\begin{array}{l}{Q}_1=129\\ Q_2=259\\ Q_3=360\end{array}$

The inter- quartile range is $231.$

For outlier

  $\begin{array}{l}{Q}_1-1.5\left(231\right)=129-346.5=-217.5\\ Q_3+1.5\left(231\right)=360+346.5=706.5\end{array}$

The lower extreme 97 and the upper extreme 694 are within the limits.

So, there are no outliers.

f.

To determine

To sketch: A box-and-whisker plot of the lengths of the rivers.

Explanation of Solution

Given information:

There are seven navigable rivers that feed into the Ohio river. The lengths of these rivers are

  

Graph:

  

Interpretation:

The vertical graph is known as box-and-whisker plots. It represents the length of the seven rivers. The median is 259. It does not have outliers.

g.

To determine

To discuss: The variability of the data.

Explanation of Solution

Given information:

There are seven navigable rivers that feed into the Ohio river. The lengths of these rivers are

  

Summary:

Variability of the data can be studied by box-and-whisker plot.

It represent the median, quartiles, interquartile range , and extreme values in a set of the data.

Terms	7
Median	259
$Q_1$	129
$Q_3$	360
Interquartile	231
Semi-interquartile	115.5
Outliers	No

Thus, the box-and-whisker drawn in previous part is a pictorial representation of the variability of the data.