Chapter 10, Problem 6CR

To determine

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

Answer to Problem 6CR

Range: $4$

Lower quartile: $6$

Median: $6.5$

Upper quartile: $8$

Interquartile range: $2$

Outliers: No outliers

Explanation of Solution

Given information:

Data set of number of fish in each fish tank: $6,5,7,8,5,6,7,9,8,6$

Calculation:

Finding range:

  $\begin{array}{l}\text{Range}=\text{highest value}-\text{lowest value}\\ =9-5\\ =4\end{array}$

Ordering the given data set:

  $5,5,6,6,6,7,7,8,8,9\to \text{in ascending order}$

Median:

  $\frac{6+7}{2}=6.5$

Lower quartile:

Median of the lower half of the data set,

  $6$

Upper quartile:

Median of the upper half of the data set,

  $8$

Interquartile range:

Difference between upper quartile and lower quartile,

  $8-6=2$

Outliers:

Multiply the interquartile range by $1.5$ ,

  $1.5\times 2=3$

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

  $\begin{array}{l}8+3=11\\ 6-3=3\end{array}$

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.