Chapter 10, Problem 8CR

To determine

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

Answer to Problem 8CR

Range: $41$

Lower quartile: $76$

Median: $82$

Upper quartile: $87.5$

Interquartile range: $11.5$

Outliers: $58$

Explanation of Solution

Given information:

Data set of Mr. Han’s students test scores:

  $99,88,81,89,77,58,92,80,83,82,74,84,76,73,99,74,82,87,82,74,86,76,85,92$

Calculation:

Finding range:

  $\begin{array}{l}\text{Range}=\text{highest value}-\text{lowest value}\\ =99-58\\ =41\end{array}$

Ordering the given data set:

  $58,73,74,74,74,76,76,77,80,81,82,82,82,83,84,85,86,87,88,89,92,92,99,99\to \text{in ascending order}$ Median:

  $82$

Lower quartile:

Median of the lower half of the data set,

  $76$

Upper quartile:

Median of the upper half of the data set,

  $87.5$

Interquartile range:

Difference between upper quartile and lower quartile,

  $87.5-76=11.5$

Outliers:

Multiply the interquartile range by $1.5$ ,

  $1.5\times 11.5=17.25$

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

  $\begin{array}{l}87.5+17.25=104.75\\ 76-17.25=58.75\end{array}$

Therefore, there is an outlier $58$ .

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