Chapter 10, Problem 7CR

To determine

To find: The measures of variability and any outliers for the given set of data.

Answer to Problem 7CR

Range: $5$

Lower quartile: $5.5$

Median: $7$

Upper quartile: $9$

Interquartile range: $3.5$

Outliers: No outliers

Explanation of Solution

Given information:

Data set of Claire’s earnings: $5,7,10,6,8$

Calculation:

Finding range:

  $\begin{array}{l}\text{Range}=\text{highest value}-\text{lowest value}\\ =10-5\\ =5\end{array}$

Ordering the given data set:

  $5,6,7,8,10\to \text{in ascending order}$

Median:

  $7$

Lower quartile:

Median of the lower half of the data set,

  $\frac{5+6}{2}=5.5$

Upper quartile:

Median of the upper half of the data set,

  $\frac{8+10}{2}=9$

Interquartile range:

Difference between upper quartile and lower quartile,

  $9-5.5=3.5$

Outliers:

Multiply the interquartile range by $1.5$ ,

  $1.5\times 3.5=5.25$

Add $3$ with the upper quartile and subtract 3 from the lower quartile,

  $\begin{array}{l}9+5.25=14.25\\ 5.5-5.25=0.25\end{array}$

Therefore, there are no outliers in the given set of data.

Thus we can find the measures of variability and any other outliers for each set of data.