Chapter 13.3, Problem 1CYU

To determine

To find: The measure of variation and any outliers for the given set of data

Answer to Problem 1CYU

Range is 145

Median is 110

Lower quartile is 83.5

Upper quartile is 145

Interquartile range is 61.5

No outliers

Explanation of Solution

Given:

The number of minutes spent bike ridings are

  $120,80,87,170,100,120,110,180,35$

Calculation:

The given data is $120,80,87,170,100,120,110,180,35$

Arrange this set of the data in numeric order

  $\text{35, 80, 87, 100, 110, 120, 120, 170, 180}$

Hence, the range is given by

  $\begin{array}{c}\text{Range }=\text{ maximum value}-\text{minimum value }\\ \text{=180}-\text{35}\\ \text{=145}\end{array}$

The median is 110

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

So, lower quartile is given by

  $\begin{array}{c}{Q}_1=\frac{80+87}{2}\\ =\frac{167}{2}\\ =83.5\end{array}$

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

So, upper quartile is given by

  $\begin{array}{c}{Q}_2=\frac{120+170}{2}\\ =\frac{290}{2}\\ =145\end{array}$

Therefore, the interquartile range is given by

  $\begin{array}{c}IQR=Q_2-Q_1\\ =145-83.5\\ =61.5\end{array}$

Now, outliers are the data which is below $Q_1-1.5\times IQR$ and above $Q_3+1.5\times IQR$

So, find the value for these expressions

  $\begin{array}{l}{Q}_1-1.5\times IQR\\ =83.5-1.5\times 61.5\\ =-8.75\end{array}$

And

  $\begin{array}{l}{Q}_3+1.5\times IQR\\ =145+1.5\times 61.5\\ =237.25\end{array}$

So, the outliers are the value which is below $-8.75$ and above $237.25$

There is no value in the given set of data in this interval.

So, there are no outliers.