Chapter 13.3, Problem 2CYP

To determine

To find: The measures of variation for the data in the table.

Answer to Problem 2CYP

Range is $28$

Median is $42$

Lower quartile is $38.5$

Upper quartile is $47$

Interquartile range is $8.5$

No outliers

Explanation of Solution

Given:

  

Calculation:

The given set of data is $40,45,49,25,44,39,53,38$

Arrange this set of the data in numeric order

  $25,38,39,40,44,45,49,53$

Hence, the range is given by

  $\begin{array}{c}\text{Range }=\text{ maximum value}-\text{minimum value }\\ \text{=53}-\text{25}\\ \text{=28}\end{array}$

The median is given by

  $\begin{array}{c}\frac{40+44}{2}\\ =\frac{84}{2}\\ =42\end{array}$

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

So, lower quartile is given by

  $\begin{array}{c}{Q}_1=\frac{38+39}{2}\\ =\frac{77}{2}\\ =38.5\end{array}$

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

So, upper quartile is given by

  $\begin{array}{c}{Q}_3=\frac{45+49}{2}\\ =\frac{94}{2}\\ =47\end{array}$

Therefore, the interquartile range is given by

  $\begin{array}{c}IQR=Q_3-Q_1\\ =47-38.5\\ =8.5\end{array}$