Chapter 13.3, Problem 11PPS

To determine

To Find:Differences in the value of the mean and the interquartile range.

Answer to Problem 11PPS

When the outlier is added, the mean and the range of the temperature is increased but the interquartile range is decreased

Explanation of Solution

Given:Find the real world data set with the at least eight values that include one or more outliers. Display the data in the table. Find the mean and the interquartile range of the data set. Then remove the outliers from the data set and find the mean and the interquartile range. Describe any differences in the value.

Let the average temperature of Maldives through year $2010-2017$

The temperature is in degree celcius.

  

Finding range Highest age = $30$

Lowest age = $14$

Then

Range = highest age − lowest age

  $\begin{array}{l}=30-14\\ =16\end{array}$

The range is $16$ .

Mean

  $\begin{array}{l}=\frac{14+14+15+16+17+19+19+30}{8}\\ =18\end{array}$

Mean= $18$

Finding Upper quartile and lower quartile of rose bowl

Arranging the data in the ascending order

  $14,14,15,16,17,19,19,30$

Median

  $\begin{array}{l}14,14,15,\underset{\_}{16,17},19,19,30\\ =\frac{16+17}{2}=16.5\end{array}$

Upper quartile

  $\begin{array}{l}14,14,15,16,17,\underset{\_}{19,19},30\\ =\frac{19+19}{2}\\ =19\end{array}$

Lower quartile

  $\begin{array}{l}14,\underset{\_}{14,15},16,17,19,19,30\\ =\frac{14+15}{2}\\ =14.5\end{array}$

For finding interquartile range Interquartile range = Upper quartile − lower quartile

  $\begin{array}{l}=19-14.5\\ =4.5\end{array}$

Now,

Multiply the interquartile range by $1.5$

We get

  $\begin{array}{l}=4.5\times 1.5\\ =6.75\end{array}$

Now subtract $6.75$ from the lower quartile and add $6.75$ to the upper quartile

I.e.

  $\begin{array}{l}14.5-6.75=7.75\\ 19+6.75=25.75\end{array}$

Thus $30$ is the only value that exist in the range of $7.75-25.75$

There is one outlier in the data which is $30$ .

Now,

After removing the outlier

Range = highest age − lowest age

  $\begin{array}{l}=19-14\\ =5\end{array}$

The range is $5$ .

Mean

  $\begin{array}{l}=\frac{14+14+15+16+17+19+19}{7}\\ =16.3\end{array}$

Mean= $16.3$

Finding Upper quartile and lower quartile of rose bowl

Arranging the data in the ascending order

  $14,14,15,16,17,19,19$

Median

  $\begin{array}{l}14,14,15,\underset{\_}{16},17,19,19\\ =16\end{array}$

Upper quartile

  $\begin{array}{l}14,14,15,16,17,\underset{\_}{19},19\\ =19\end{array}$

Lower quartile

  $\begin{array}{l}14,\underset{\_}{14},15,16,17,19,19\\ =14\end{array}$

For finding interquartile range Interquartile range = Upper quartile − lower quartile

  $\begin{array}{l}=19-14\\ =5\end{array}$

Hence,

When the outlier is added, the mean and the range of the temperature is increased but the interquartile range is decreased.