Chapter 13.3, Problem 5GP

(a)

To determine

To Compare: Homeroom $104$ range with Homeroom $102$ . Does either Homeroom have outlier in the data set.

Answer to Problem 5GP

Homeroom $104$ has higher range than Homeroom $102$

The only outlier in the data of Homeroom $104$ is $51$

Explanation of Solution

Given:The Number of magazines sold by each student Homeroom $102$ and Homeroom $104$ is shown in the stem and leaf plot.

  

First we finding the range of Homeroom $102$

Range = Highest number − lowest number

  $\begin{array}{l}=40-2\\ =38\end{array}$

Thus

The range of Homeroom $102$ is $38$

Finding the range of Homeroom $104$

Range = Highest number − lowest number

  $\begin{array}{l}=51-1\\ =50\end{array}$

Thus

The range of Homeroom $104$ is $50$

Now

Finding median, upper quartile, lower quartile of Homeroom $102$

Arranging the data in ascending order. After arranging the data we can find the median, upper quartile, lower quartile

Thus

  $2,5,7,7,8,9,13,14,,14,16,16,23,24,28,33,35,35,40$

Median

  $\begin{array}{l}2,5,7,7,8,9,13,14,\underset{\_}{14,16},16,23,24,28,33,35,35,40\\ median=\frac{14+16}{2}=15\end{array}$

Lower quartile

  $\begin{array}{l}2,5,7,7,\underset{\_}{8},9,13,14,,14,16,16,23,24,28,33,35,35,40\\ \text{lower Quartile = 8}\end{array}$

Upper quartile

  $\begin{array}{l}2,5,7,7,8,9,13,14,,14,16,16,23,24,\underset{\_}{28},33,35,35,40\\ \text{Upper quartile = 28}\end{array}$

Now finding interquartile range

Interquartile range = Upper quartile − Lower quartile

  $\begin{array}{l}=28-8\\ =20\text{ magazines}\end{array}$

Now, after finding the inter quartile range, multiply it by $1.5$

  $\begin{array}{l}=20\times 1.5\\ =30\end{array}$

Finally subtract $15$ from the lower quartile and add $15$ to the upper quartile

i.e.

  $\begin{array}{l}15-30=-15\\ 28+30=58\end{array}$

It is concluded that there are no outliers in the data of the homeroom $102$

Finding median, upper quartile, lower quartile of Homeroom $104$

Arranging the data in ascending order. After arranging the data we can find the median, upper quartile, lower quartile

Thus

  $1,2,4,8,9,9,10,10,11,11,12,15,16,16,20,25,26,51$

Median

  $\begin{array}{l}1,2,4,8,9,9,10,10,\underset{\_}{11,11},12,15,16,16,20,25,26,51\\ median=\frac{11+11}{2}=11\end{array}$

Lower quartile

  $\begin{array}{l}1,2,4,8,\underset{\_}{9},9,10,10,11,11,12,15,16,16,20,25,26,51\\ \text{lower Quartile = 9}\end{array}$

Upper quartile

  $\begin{array}{l}1,2,4,8,9,9,10,10,11,11,12,15,16,\underset{\_}{16,}20,25,26,51\\ \text{Upper quartile =16}\end{array}$

Now finding interquartile range

Interquartile range = Upper quartile − Lower quartile

  $\begin{array}{l}=16-9\\ =7\text{ magazines}\end{array}$

Now, after finding the inter quartile range, multiply it by $1.5$

  $\begin{array}{l}=7\times 1.5\\ =10.5\end{array}$

Finally subtract $10.5$ from the lower quartile and add $10.5$ to the upper quartile

i.e.

  $\begin{array}{l}9-10.5=-1.5\\ 16+10.5=26.5\end{array}$

It is concluded that the only outlier in the data of Homeroom $104$ is $51$

Hence, Homeroom $104$ has higher range than Homeroom $102$

The only outlier in the data of Homeroom $104$ is $51$

(b)

To determine

To Explain:How does the outlier affect the range for the number of magazines sold in that homeroom?

Answer to Problem 5GP

When the outlier is included, the range doubled the number of magazines in the homeroom $104$

Explanation of Solution

Given: The Number of magazines sold by each student Homeroom $102$ and Homeroom $104$ is shown in the stem and leaf plot.

  

First we have to calculate the range of Homeroom $104$ with and without the outlier

Range without outlier

  $\begin{array}{l}=26-1\\ =25\end{array}$

Range with the outlier

  $\begin{array}{l}=51-1\\ =50\end{array}$

It is concluded that when the outlier is included, the range doubled the number of magazines in the homeroom $104$