#### Solved Examples and Worksheet for Identifying Outliers

Q1The loss percentages of a company during the last 5 years is represented in the bar graph below. What is the outlier loss percentage and in which year did it occur? A. 6%, 1997
B. 2%, 1997
C. 1%, 1998
D. 5%, 1999

Step: 1
Outlier of a data set is the data value that is far apart from the remaining values.
Step: 2
The height of each bar in the graph indicate the percentage loss of the company during the particular year.
Step: 3
From the graph, the loss percentages in the last 5 years are 5%, 4%, 6%, 1% and 5%.
Step: 4
Among the values 1% is far apart from the remaining ones.
Step: 5
So, the outlier loss percentage is 1%, which occured in the year 1998.
Correct Answer is :   1%, 1998
Q2The rates of cheese pizzas at different pizza shops are $3,$2.75, $3.50,$1, $2,$2.75, $3.50. Find the outlier of the rates. A.$2
B. $3.50 C.$1
D. No outlier

Step: 1
Outlier is the data value which is far apart from the remaining values.
Step: 2
In the data, there is no value which is far apart from the rest of the values.
[All values are in the same range.]
Step: 3
So, there is no outlier for the data.
Correct Answer is :   No outlier
Q3The number of students in different grades of a school are 21, 25, 22, 26 and 20. Find the outlier of the number of students.
A. 22
B. 21
C. No outlier
D. 26

Step: 1
An outlier is the data item which is far apart from the remaining values.
Step: 2
No data value among the values is far apart from the others.
Step: 3
So, there is no outlier for the data set.
Correct Answer is :   No outlier
Q4William has taken an aptitude test 8 times and his scores are 96, 98, 98, 105, 36, 87, 95 and 93. Is mean or median the better measure of central tendency?

A. Median
B. Mean

Step: 1
Mean is a better measure of central tendency if there is no outlier for the data.
Step: 2
36 is the outlier of the data as it is far apart from other data values.
Step: 3
So, it may skew the central tendency.
Step: 4
As the outlier influences the mean, the median is the better measure of the central tendency.
Q5What is the outlier in the box-and-whisker plot? A. 65
B. 50
C. 45
D. 10

Step: 1
The outlier in a box-and-whisker plot is the data item, which is much higher or much lower than the other set of data items.
Step: 2
In the plot, the number 10 is much farther from the remaining set of data items.
Step: 3
So, the outlier in the plot is 10.
Q6John has taken an aptitude test 6 times and his scores are 82, 85, 90, 95, 27 and 87. Which measure of central tendency is most appropriate for his scores?
A. Median
B. Range
C. Mode
D. Mean

Step: 1
Range is the difference between the maximum and minimum values of the data set.
Step: 2
Mode is used to find the occurrence of maximum times of a data value.
Step: 3
Mean is a better measure of central tendency if there is no outlier in the data.
Step: 4
27 is the outlier in the data as it is far apart from other values. So, it may skew the central tendency.
Step: 5
As the outlier influences the mean, the median is the better measure of the central tendency.
Q7Lauren has taken a math test 10 times and her scores are 95, 99, 99, 98, 45, 88, 93, 95, 86 and 96. Which measure of central tendency is the most appropriate for her scores?

A. Median
B. Mean
C. Mode
D. Range

Step: 1
Mean is a better measure of central tendency if there is no outlier for the data.
Step: 2
45 is the outlier of the data as it is far apart from other data values.
Step: 3
So, it may skew the central tendency.
Step: 4
As the outlier influences the mean, the median is the better measure of the central tendency.
Q8The different weights of students in a class are given as 36, 40, 45, 76, and 29. Check for outlier.

A. 76
B. 29
C. 60.5
D. no outlier

Step: 1
Arrange the weights of students in increasing order.
29, 36, 40, 45, 76
Step: 2
Q2 = 40
[Q2 is the median, which is the middle value of the data set in the order.]
Step: 3
Q1 = 29 + 362 = 32.5
[Q1 is the median of the data values less than Q2.]
Step: 4
Q3 = 45 + 762 = 60.5
[Q3 is the median of the data values greater than Q2.]
Step: 5
Interquartile Range (IQR) = Q3 - Q1
[Formula.]
Step: 6
= 60.5 - 32.5 = 28
Step: 7
To find the outlier, compute the cut-off points for outliers.
Step: 8
Lower fence = Q1 - 1.5(IQR)
= 32.5 - 1.5(28) = - 9.5
[Formula.]
Step: 9
Upper fence = Q3 +1.5(IQR)
= 60.5 + 1.5(28) = 102.5
[Formula]
Step: 10
If a data value is less than the lower fence or greater than the upper fence, then it is considered as an outlier.
Step: 11
There are no values that are less than - 9.5 or greater than 102.5.
Step: 12
So, there is no outlier.
Correct Answer is :   no outlier
Q9Check the following data set for outliers.
3.5, 4.7, 5.3, 6.8, 4.2, 5.8, 12.5

A. 2.6
B. 5.3
C. no outlier
D. 12.5

Step: 1
Arrange the data values in increasing order.
3.5, 4.2, 4.7, 5.3, 5.8, 6.8, 12.5
Step: 2
Q2 = 5.3
[Q2 is the median, which is the middle value of the data set in the order.]
Step: 3
Q1 = 4.2
[Q1 is the median of the data values less than Q2.]
Step: 4
Q3 = 6.8
[Q3 is the median of the data values greater than Q2.]
Step: 5
Interquartile Range (IQR) = Q3 - Q1
= 6.8 - 4.2 = 2.6
[Formula.]
Step: 6
To find the outlier, compute the cut-off points for outliers.
Step: 7
Lower fence = Q1 - 1.5(IQR)
= 4.2 - 1.5(2.6) = 0.3
[Formula.]
Step: 8
Upper fence = Q3 + 1.5(IQR)
= 6.8 + 1.5(2.6) = 10.7
Step: 9
If a data value is less than the lower fence or greater than the upper fence, then it is considered as an outlier.
Step: 10
The value 12.5 is greater than 10.7.
Step: 11
So, 12.5 is the outlier.
Q10The number of days taken by a company to manufacture different sizes of same product is given as 7, 8, 23, 17, 15, 19, 12, and 45. Check for outliers.
A. 45
B. 7
C. 19
D. no outlier

Step: 1
Arrange the number of days taken in increasing order.
7, 8, 12, 15, 17, 19, 23, 45
Step: 2
Q2 = 15 + 172 = 16
[Q2 is the median, which is the mean of the middle values of the data set in the order.]
Step: 3
Q1 = 8 + 122 = 10
[Q1 is the median of the data values less than Q2.]
Step: 4
Q3 = 19 + 232 = 21
[Q3 is the median of the data values greater than Q2.]
Step: 5
Interquartile Range(IQR) = Q3 - Q1
= 21 - 10 = 11
[Formula.]
Step: 6
To find the outlier, compute the cut-off points for outliers.
Step: 7
Lower fence = Q1 - 1.5(IQR)
= 10 - 1.5(11) = - 6.5
[Formula]
Step: 8
Upper fence = Q3 + 1.5(IQR)
= 21 + 1.5(11) = 37.5
[Formula]
Step: 9
If a data value is less than the lower fence or greater than the upper fence, then it is considered as an outlier.
Step: 10
The value 45 is greater than 37.5.
Step: 11
So, 45 is the outlier.