Chapter 13.3, Problem 6PPS

To determine

To Find:Measures of variation and any outliers for each set of data.

Answer to Problem 6PPS

Median= $0.56$

Lower quartile = $0.46$

Upper quartile = $0.63$

Interquartile= $0.17$

Outlier= $1.25$

Explanation of Solution

Given: Find the measures of variation and any outliers for each set of data.

  

Highest vale = $1.25$

Lowest value = $0.2$

Range = Highest value − lowest value

  $\begin{array}{l}=1.25-0.2\\ =1.05\end{array}$

For median:

Median

  $\begin{array}{l}=\frac{0.53+0.58}{2}\\ =0.56\end{array}$

Then

Lower quartile = $0.46$

Upper quartile = $0.63$

Interquartile= Upper quartile − lower quartile

  $\begin{array}{l}=0.63-0.46\\ =0.17\end{array}$

Multiply the interquartile range by $1.5$

  $0.17\times 1.5=0.255$

Subtract $0.255$ from the lower quartile and add $0.255$ to the upper quartile

  $\begin{array}{l}0.46-0.255=0.205\\ 0.63+0.255=0.88\\ Outlier=1.25\end{array}$

Hence there is one outliers= $1.25$