Chapter 14.3, Problem 23E

a.

To determine

To calculate: The quartile points of the data.

Answer to Problem 23E

The quartile points are $Q_1=424,Q_2=538,\text{ and }{Q}_3=707.5$

Explanation of Solution

Given information:

The data is on the college tution and fees.

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 ,

  $\left\{\$320,\$397,\$451,\$520,\$556,\$648,\$767,\$967\right\}.$

So, $n=8$ .

There is even number of data.

The median of even number of data is −

  $\begin{array}{l}=\frac{{\left(\frac{8}{2}\right)}^{th}\text{term}+{\left(\frac{8}{2}+1\right)}^{th}\text{term}}{2}\\ =\frac{{\left(4\right)}^{th}\text{term}+{\left(5\right)}^{th}\text{term}}{2}\\ =\frac{520+556}{2}\\ =\frac{1076}{2}\\ =538\end{array}$

Therefore , the data is divided into two parts.

Lower half - $\$320,\$397,\$451,\$520$ and Upper half - $\$556,\$648,\$767,\$967.$

Median of lower half is −

  $\begin{array}{l}=\frac{{\left(\frac{4}{2}\right)}^{th}\text{term}+{\left(\frac{4}{2}+1\right)}^{th}\text{term}}{2}\\ =\frac{{\left(2\right)}^{nd}\text{term}+{\left(3\right)}^{rd}\text{term}}{2}\\ =\frac{397+451}{2}\\ =\frac{848}{2}\\ =424\end{array}$

Median of upper half is −

  $\begin{array}{l}=\frac{{\left(\frac{4}{2}\right)}^{th}\text{term}+{\left(\frac{4}{2}+1\right)}^{th}\text{term}}{2}\\ =\frac{{\left(2\right)}^{nd}\text{term}+{\left(3\right)}^{rd}\text{term}}{2}\\ =\frac{648+767}{2}\\ =\frac{1415}{2}\\ =707.5\end{array}$

The quartile points are :-

  $\begin{array}{l}{Q}_1=424\\ Q_2=538\\ Q_3=707.5\end{array}$

Hence, the quartile points are $Q_1=424,Q_2=538,\text{ and }{Q}_3=707.5$

b.

To determine

To calculate: The interquartile range .

Answer to Problem 23E

The interquartile range is $\mathrm{283.5.}$

Explanation of Solution

Given information:

The data is on the college tution and fees.

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 ,

  $\left\{\$320,\$397,\$451,\$520,\$556,\$648,\$767,\$967\right\}.$

So, $n=8$ .

There is even number of data.

The median of even number of data is $538.$

The quartile points are :-

  $\begin{array}{l}{Q}_1=424\\ Q_2=538\\ Q_3=707.5\end{array}$

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

  $\begin{array}{l}=707.5-424\\ =\mathrm{283.5.}\end{array}$

Hence, the interquartile range is $\mathrm{283.5.}$

c.

To determine

To show: whether there is any outliers.

Answer to Problem 23E

No outliers.

Explanation of Solution

Given information:

The data is on the college tution and fees.

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 ,

  $\left\{\$320,\$397,\$451,\$520,\$556,\$648,\$767,\$967\right\}.$

There is even number of data.

The median of even number of data is $538.$

The quartile points are :-

  $\begin{array}{l}{Q}_1=424\\ Q_2=538\\ Q_3=707.5\end{array}$

The inter- quartile range is $\mathrm{283.5.}$

For outlier

$\begin{array}{l}{Q}_1-1.5\left(283.5\right)=424-425.25=-1.25\\ Q_3+1.5\left(283.5\right)=707.5+425.25=1132.75\end{array}$

The lower extreme 320 and the upper extreme 967 are within the limits.

So, there are no outliers.

d.

To determine

To sketch: A box-and-whisker of the number of women’s teams playing a sport.

Explanation of Solution

Given information:

The data is on the college tution and fees.

Graph:

  

Interpretation:

The horizontal graph is known as box-and-whisker plots. It represents the data on the college tuition and fees. The median is 538. It does not have outliers.

e.

To determine

To calculate: The mean deviation of the data.

Answer to Problem 23E

The mean deviation is $169.53125$

Explanation of Solution

Given information:

The data is on the college tution and fees.

Formula used:

The arithmetic mean of the absolute values of the deviations from the mean of a set of data is called the mean deviation.

  $MD=\frac{1}{n}{\displaystyle \sum _{i=1}^n\left|X_i-\overline{X}\right|}$

Formula of Mean

  $\overline{X}=\frac{{\displaystyle \sum X}}{n}$ where X represents each of the values in the data set.

Calculation:

Consider the data in ascending order ,

  $\left\{\$320,\$397,\$451,\$520,\$556,\$648,\$767,\$967\right\}.$

So, $n=8$ .

There is even number of data.

The mean of the data is

  $\begin{array}{l}=\frac{320+397+451+520+556+648+767+967}{8}\\ =578.25\end{array}$

The mean deviation is −

$X_i$	$\overline{X}$	$\left|X_i-\overline{X}\right|$
320	578.25	$\begin{array}{l}=\left|320-578.25\right|\\ =320\end{array}$
397	578.25	$\begin{array}{l}=\left|397-578.25\right|\\ =181.25\end{array}$
451	578.25	$\begin{array}{l}=\left|451-578.25\right|\\ =127.25\end{array}$
520	578.25	$\begin{array}{l}=\left|520-578.25\right|\\ =58.25\end{array}$
556	578.25	$\begin{array}{l}=\left|556-578.25\right|\\ =22.25\end{array}$
648	578.25	$\begin{array}{l}=\left|648-578.25\right|\\ =69.75\end{array}$
767	578.25	$\begin{array}{l}=\left|767-578.25\right|\\ =188.75\end{array}$
967	578.25	$\begin{array}{l}=\left|967-578.25\right|\\ =388.75\end{array}$
${\displaystyle \sum _{i=1}^8\left|X_i-\overline{X}\right|}$		$1356.25$$$

  $\begin{array}{l}MD=\frac{1}{8}\left(1356.25\right)\\ =169.53125\end{array}$

Hence, the mean deviation is $169.53125$

f.

To determine

To calculate: The standard deviation of the data.

Answer to Problem 23E

The standard deviation is $97461.55$

Explanation of Solution

Given information:

The data is on the college tution and fees.

Formula used:

A measure of variability associated with the arithmetic mean is the standard deviation.

  $\sigma =\sqrt{\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}$

Calculation:

Consider the data in ascending order ,

  $\left\{\$320,\$397,\$451,\$520,\$556,\$648,\$767,\$967\right\}.$

So, $n=8$ .

There is even number of data.

The mean of the data is

  $\begin{array}{l}=\frac{320+397+451+520+556+648+767+967}{8}\\ =578.25\end{array}$

The standard deviation is −

$X_i$	$\overline{X}$	$\left(X_i-\overline{X}\right)$	${\left(X_i-\overline{X}\right)}^2$
320	578.25	$\begin{array}{l}=320-578.25\\ =-320\end{array}$	$102400$
397	578.25	$\begin{array}{l}=397-578.25\\ =-181.25\end{array}$	$32851.5625$
451	578.25	$\begin{array}{l}=451-578.25\\ =-127.25\end{array}$	$16192.5625$
520	578.25	$\begin{array}{l}=520-578.25\\ =-58.25\end{array}$	$3393.0625$
556	578.25	$\begin{array}{l}=556-578.25\\ =-22.25\end{array}$	$495.0625$
648	578.25	$\begin{array}{l}=648-578.25\\ =69.75\end{array}$	$4865.0625$
767	578.25	$\begin{array}{l}=767-578.25\\ =188.75\end{array}$	$35626.5625$
967	578.25	$\begin{array}{l}=967-578.25\\ =388.75\end{array}$	$151126.5625$
${\displaystyle \sum _{i=1}^8\left|X_i-\overline{X}\right|}$$$			$346950.4375$

  $\begin{array}{l}=\sigma =\sqrt{\frac{1}{8}\left(346950.4375\right)}\\ =\sigma =\sqrt{43368.80}\\ =\sigma =208.251\end{array}$

Hence, the standard deviation is $\mathrm{208.251.}$

g.

To determine

To discuss: The variability of the data.

Explanation of Solution

Given information:

The data is on the college tution and fees.

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	538
$Q_1$	424
$Q_3$	707.5
Interquartile	283.5
Semi-interquartile	141.75
Outliers	No

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