If a value in a data set is more than 1.5 times the away from the lower or upper , it is considered an . This is why, to identify any outliers, we first have to find these statistical measures, including any outliers.
Analyzing the Data With Any Outliers
We want to find the , , and of the given .
85, 77, 211, 88, 91, 84, 85
Let's begin by calculating the mean.
Mean
The mean of a data set is the of the values by the total number of values in the set. Let's start by calculating the sum of the given values.
85+ 77+ 211+ 88+ 91+ 84+ 85
= 721
There are 7 values in our set, so we have to divide the sum by 7.
Mean: 721/7=103
The mean is 103. We can continue by finding the median.
Median
When the data are arranged in numerical order, the
median is the middle value — or the mean of the two middle values — in a set of data. Let's arrange the given values and find the median.
77, 84, 85, 85, 88, 91, 211
The median is 85.
Median : 85
The last measure we need is the mode. Let's find it!
Mode
The mode is the value or values that appear most often in a set of data. Arranging the data set from least to greatest makes it easier to see how often each value appears. Let's arrange the values before we find the mode.
77, 84, 85, 85, 88, 91, 211
We can see that the most common value in the given data set is 85. This is the mode of our data set.
Mode: 85
Identifying Outliers
To identify any outliers, we have to calculate the interquartile range (IQR). To do this let's recall some information about the quartiles first!
- Second Quartile (Q_2) is the median of the data set. It divides the set of data into two halves.
- Lower Quartile (Q_1) is the median of the lower half of the data set.
- Upper Quartile (Q_3) is the median of the upper half of the data set.
- Interquartile Range is the difference between the and the (Q_3-Q_1).
Let's start by recalling the ordered data set from least to greatest value!
77, 84, 85, 85, 88, 91, 211
The median of the set is 85. This value divides the set into two halves. We have one middle values for each half.
Upper Quartile:& 91
Lower Quartile:& 84
The next step to calculate the interquartile range is to calculate the difference between the upper and lower quartiles. Let's do it!
Interquartile Range:& 91- 84= 7
Next, we need to determine the maximum and minimum values for data to be considered an outlier. Outliers are more than 1.5 times the IQR away from the upper and lower quartiles. Let's break it into two steps and start with the minimum value.
Outlier < Q_1-1.5*IQR
Let's substitute 84 for Q_1 and 7 for IQR.
Q_1 - 1.5 * IQR
84 - 1.5 * 7
84 - 10.5
73.5
Q_2 + 1.5 * IQR
91 + 1.5 * 7
91 + 10.5
101.5
Analyzing the Data Without the Outliers
Let's repeat the process, this time excluding 211 from the data set.
Mean
Let's start by calculating the sum of the given values without 211.
77+ 84+ 85+ 85+ 88+ 91
= 510
There are 6 values in our set, because we excluded an outlier — 211.
Mean: 510/6=85
Median
Recall the ordered set with 211 excluded.
77, 84, 85, 85, 88, 91
This time the number of values in our set is 6. This is why the median is the mean of the two middle values.
Median : 85+ 85/2 = 85
Mode
Let's take a look at the ordered set with outlier excluded one more time!
77, 84, 85, 85, 88, 91
We can see that the most common value in the given data set is still 85. This is the mode of our data set.
Mode: 85
Summary
Finally, we summarize our findings in the table below so it is easier to compare the results.
| Data set
|
Mean
|
Median
|
Mode
|
| With Outlier
|
103
|
85
|
85
|
| Without Outlier (211)
|
85
|
85
|
85
|
We can see that removing the outlier modified only the mean. The mean without outlier decreased. This is because the outlier was excluded with the maximum value.
