Chapter 10, Problem 5CR

To determine

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

Answer to Problem 5CR

Range: $4$

Lower quartile: $32$

Median: $33$

Upper quartile: $34$

Interquartile range: $2$

Outliers: No outliers

Explanation of Solution

Given information:

Data set of number of minutes spent reading each night: $31,33,32,34,35,33$

Calculation:

Finding range:

  $\begin{array}{l}\text{Range}=\text{highest value}-\text{lowest value}\\ =35-31\\ =4\end{array}$

Ordering the given data set:

  $31,32,33,33,34,35\to \text{in ascending order}$

Median:

  $\frac{33+33}{2}=33$

Lower quartile:

Median of the lower half of the data set,

  $32$

Upper quartile:

Median of the upper half of the data set,

  $34$

Interquartile range:

Difference between upper quartile and lower quartile,

  $34-32=2$

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,

  $\begin{array}{l}34+3=37\\ 32-3=29\end{array}$

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.