Chapter 1.3, Problem 107E

To determine

To identify the outliers if any.

Answer to Problem 107E

Interquartile Range = 2.8

Explanation of Solution

Given:

Mean = 3.2

Standard deviation = 1.9

  

Formula used:

  $\begin{array}{l}{\text{Outliers > Q}}_{\text{3}}+1.5\text{IQR}\\ {\text{Outliers < Q}}_1-1.5\text{IQR}\end{array}$

Calculation:

From the question 105:

First quartile = 1.9

Third quartile = 4.7

Interquartile Range: 2.8

Now,

  $\begin{array}{l}{\text{Outliers > Q}}_{\text{3}}+1.5\text{IQR}\\ \text{Outliers > 4}\text{.7 + 1}\text{.5 }\times \text{2}\text{.8}\\ \text{Outliers > 8}\text{.9}\\ \\ {\text{Outliers < Q}}_1-1.5\text{IQR}\\ \text{Outliers <1}\text{.9 - 1}\text{.5}\times \text{2}\text{.8}\\ \text{Outliers < -2}\text{.3}\end{array}$

It shows that any value that away from the range of -2.3 to 8.9, will be considered as the outlier and based on the given dot plot, it can easily observed that no value falls beyond this range.