Chapter 13.3, Problem 3CYP

To determine

To find: Any outliers in the data set.

Answer to Problem 3CYP

  $155$

Explanation of Solution

Given:

  

Calculation:

The given set of data is $105,120,155,115,96,100,110,120$

Arrange this set of the data in numeric order

  $96,100,105,110,115,120,120,155$

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

So, lower quartile is given by

  $\begin{array}{c}{Q}_1=\frac{100+105}{2}\\ =\frac{205}{2}\\ =102.5\end{array}$

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

So, upper quartile is given by

  $\begin{array}{c}{Q}_3=\frac{120+120}{2}\\ =\frac{240}{2}\\ =120\end{array}$

Therefore, the interquartile range is given by

  $\begin{array}{c}IQR=Q_3-Q_1\\ =120-102.5\\ =17.5\end{array}$

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

  $\begin{array}{l}{Q}_1-1.5\times IQR\\ =102.5-1.5\times 17.5\\ =76.25\end{array}$

And

  $\begin{array}{l}{Q}_3+1.5\times IQR\\ =120+1.5\times 17.5\\ =146.25\end{array}$

So, the outliers are the value which is below $76.25$ and above $146.25$

Only $155$ lie in this interval.

Hence, the outlier for the given set of data is $155$