Chapter 13.3, Problem 20STP

To determine

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

Answer to Problem 20STP

Range is $68$

Median is $36.5$

Lower quartile is $30$

Upper quartile is $53$

Interquartile range is $23$

No outlier

Explanation of Solution

Given:

  $\text{30,62,35,80,12,24,30,39,53,38}$

Calculation:

The given data is $\text{30,62,35,80,12,24,30,39,53,38}$

Arrange this set of the data in numeric order

  $\text{12, 24, 30, 30, 35, 38, 39, 53, 62, 80}$

Hence, the range is given by

  $\begin{array}{c}\text{Range }=\text{ maximum value}-\text{minimum value }\\ =80-12\\ =68\end{array}$

The median is given by

  $\begin{array}{l}\frac{35+38}{2}\\ =\frac{73}{2}\\ =36.5\end{array}$

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

So, lower quartile is given by

  $Q_1=30$

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

So, upper quartile is given by

  $Q_3=53$

Therefore, the interquartile range is given by

  $\begin{array}{c}IQR=Q_3-Q_1\\ =53-30\\ =23\end{array}$

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

  $\begin{array}{l}{Q}_1-1.5\times IQR\\ =30-1.5\times 23\\ =-4.5\end{array}$

And

  $\begin{array}{l}{Q}_3+1.5\times IQR\\ =53+1.5\times 23\\ =87.5\end{array}$

So, the outliers are the value which is below $-4.5$ and above $87.5$

No value in the given set of data lies in this interval.

Hence, there is no outlier.