Chapter 13.3, Problem 4PPS

To determine

To find: The measure of variation and any outliers for the given set of data

Answer to Problem 4PPS

Range is $70$

Median is $92$

Lower quartile is $89$

Upper quartile is $104$

Interquartile range is $15$

Outlier is $150$

Explanation of Solution

Given:

  

Calculation:

The given data is $89,90,80,100,92,104,150$

Arrange this set of the data in numeric order

  $80,89,90,92,100,104,150$

Hence, the range is given by

  $\begin{array}{c}\text{Range }=\text{ maximum value}-\text{minimum value }\\ \text{=}150-80\\ =70\end{array}$

The median is 92

The lower quartile is the median of the left half of the data.

So, lower quartile is given by

  $Q_1=89$

The upper quartile is the median of the right half of the data.

So, upper quartile is given by

  $Q_3=104$

Therefore, the interquartile range is given by

  $\begin{array}{c}IQR=Q_3-Q_1\\ =104-89\\ =15\end{array}$

Now, outliers are the data which is below $Q_1-1.5\times IQR$ and above $Q_3+1.5\times IQR$

So, find the value for these expressions

  $\begin{array}{l}{Q}_1-1.5\times IQR\\ =89-1.5\times 15\\ =66.5\end{array}$

And

  $\begin{array}{l}{Q}_3+1.5\times IQR\\ =104+1.5\times 15\\ =126.5\end{array}$

So, the outliers are the value which is below $66.5$ and above $126.5$

So, the outlier for the given set of data is $150$