Question

check the following data set for outliers. if there is more than one answer, separate them with commas. select
one\ if applicable.
132, 136, 110, 135, 141, 146, 153, 160, 143, 142
send data to excel
outlier(s):
none
undo

Explanation:

Step1: Order the data set

First, we order the data set from smallest to largest: \(110, 132, 135, 136, 141, 142, 143, 146, 153, 160\)

Step2: Find the median (Q2)

Since there are 10 data points (even number), the median is the average of the 5th and 6th values.
The 5th value is \(141\) and the 6th value is \(142\). So, \(Q2=\frac{141 + 142}{2}=\frac{283}{2} = 141.5\)

Step3: Find Q1 (median of lower half)

The lower half of the data is \(110, 132, 135, 136, 141\) (first 5 values). The median of this lower half is the 3rd value, so \(Q1 = 135\)

Step4: Find Q3 (median of upper half)

The upper half of the data is \(142, 143, 146, 153, 160\) (last 5 values). The median of this upper half is the 3rd value, so \(Q3=146\)

Step5: Calculate the interquartile range (IQR)

\(IQR=Q3 - Q1=146 - 135 = 11\)

Step6: Determine the lower and upper fences

The lower fence is \(Q1-1.5\times IQR=135-1.5\times11 = 135 - 16.5=118.5\)
The upper fence is \(Q3 + 1.5\times IQR=146+1.5\times11=146 + 16.5 = 162.5\)

Step7: Identify outliers

We check each data point to see if it is below the lower fence or above the upper fence.

• \(110\) is less than \(118.5\) (lower fence), so it is an outlier.
• All other data points (\(132, 135, 136, 141, 142, 143, 146, 153, 160\)) are between \(118.5\) and \(162.5\), so they are not outliers.

Answer:

\(110\)