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 .
101, 110, 99, 100, 64, 112, 110, 111, 102
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.
101+ 110+ 99+ 100+ 64+ 112+ 110+ 111+ 102
= 909
There are 9 values in our set, so we have to divide the sum by 9.
Mean: 909/9=101
The mean is 101. 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.
64, 99, 100, 101, 102, 110, 110, 111, 112
The number of values in our set is 9. This means that there is one middle value. This is why the median is 102.
Median : 102
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.
64, 99, 100, 101, 102, 110, 110, 111, 112
We can see that the most common value in the given data set is 110. This is the mode of our data set.
Mode: 110
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!
64, 99, 100, 101, 102, 110, 110, 111, 112
The median of the set is 102. This value divides the set into two halves. We have two middle values for each half. Thus, we need to calculate the mean of those middle values.
Upper Quartile:& 110+ 111/2=110.5
Lower Quartile:& 99+ 100/2= 99.5
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:& 110.5- 99.5= 11
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 99.5 for Q_1 and 11 for IQR.
Q_1 - 1.5 * IQR
99.5 - 1.5 * 11
99.5 - 16.5
83
Q_2 + 1.5 * IQR
110.5 + 1.5 * 11
110.5 + 16.5
127
Analyzing the Data Without the Outliers
Let's repeat the process, this time excluding 64 from the data set.
Mean
Let's start by calculating the sum of the given values without 64.
99+ 100+ 101+ 102+110+ 110+ 111+ 112
= 845
There are 8 values in our set, because we excluded an outlier — 64.
Mean: 845/8=105.625
Median
Recall the ordered set with 64 excluded.
99, 100, 101, 102, 110,110, 111, 112
This time the number of values in our set is 8. This is why there are 2 middle values. The median is the mean of them.
Median : 102+ 110/2= 106
Mode
The most repeated value is stil 110. The mode of the set remains the same.
Mode: 110
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
|
101
|
102
|
110
|
| Without Outlier (64)
|
105.625
|
106
|
110
|
We can see that removing the outlier modified the mean and the median. They both increased but the mean increased more than the median. This is because the outlier was excluded with the minimum value. This set without outlier still has the same mode.
