Chapter 1.3, Problem 125E

To determine

To find:The outliers of the stem plot graph.

Answer to Problem 125E

The outlier is 216

Explanation of Solution

Given:

As per the given graph,the given dataset is:

91	95	103	107	108	109	114	117	121	129
148	149	153	154	154	154	155	162	163	169
172	172	173	183	185	186	191	203	210	222

Calculation:

Thus, the total number of values (n) is 30.

First quartile (Q₁), could be calculated as:

  $\begin{array}{c}{Q}_1=\frac{{\left(n+1\right)}^{th}term}{4}\\ ={\frac{\left(30+1\right)}{4}}^{th}term\\ ={7.75}^{th}term\\ =7^{th}term+\frac{3}{4}\left(8^{th}-7^{th}\right)term\\ =114+\frac{3}{4}\left(117-114\right)\\ =115.80\end{array}$

Thus, first Quartile is 115.80.

Third quartile (Q₃), can be calculated as:

  $\begin{array}{c}{Q}_3={\left(\frac{3\left(n+1\right)}{4}\right)}^{th}term\\ ={\left(\frac{3\left(31+1\right)}{4}\right)}^{th}term\\ ={24}^{th}term\\ =183\end{array}$

Thus, the third quartile is 22.

The upper and lower limit for the value can be calculated as:

  $\begin{array}{c}\text{Lower value limit}\left(LL\right)=Q_1-1.5\left(Q_3-Q_1\right)\\ =115.80-1.5\left(183-115.80\right)\\ =15\\ \text{Upper value limit}\left(UL\right)=Q_3+1.5\left(Q_3-Q_1\right)\\ =115.80+1.5\left(183-115.80\right)\\ =216.6\end{array}$

Thus, those value which does not lie in the interval (15,216.6), would be outlier.

Since, the values 216does not lie in the interval.

Hence, the outlier is 216.

Thus, the correct option is “c”.