Measures of Central Tendency
Measures of Central Tendency
The Median
Median:
The median is an alternative measure to the mean. It is the value of the
middle item in the set.
50% of the data lie on either side of the value
Relatively easy to find
Can be used with open ended data
Is not influenced by extreme values
Easy to understand
Cannot be used for further analysis
To find the median:
Case 1: When the data is a simple list of numbers arranged in ascending order, then
the median is the middle value (if there is an odd number of values) or the average of
the middle two values (if there is an even number of values)
Example:
43,47,48, 50, 51, 51, 75
median = 50
1,2,2,2,3,3,4,6
median = (3+2)/2 = 2.5
Case 2: When the data is arranged in a frequency distribution the following formula is
used:
1. Find the Median Class i.e. the interval in which the middle item (N/2) lies.
2. Median:
L +
(N 2 −Cumulative frequency before median Class )
.Class width
(Frequency of Median Class )
Where L = the lower boundary of the median interval
And N = the total number of items.
An alternative way of writing down this formula is:
⎡N
⎤
⎢ 2 − FM −1 ⎥
Median = Lm + ⎢
⎥cm
fm
⎢
⎥
⎣⎢
⎦⎥
where: Lm
= lower bound of the median class
FM-1 = the total of all the classes before the median class
fm
cm
= actual frequency of the median class
= median class width (the class interval)
Walter Fleming
Page 1 of 4
Measures of Central Tendency
Example: Find the median age of the representatives:
Age
Number of
representatives
20 - 25
2
Totals to end
of each class
25 - 30
14
2
35 - 40
43
30 - 35
29
16
45
40 -45
33
88
Over 45
9
121
130
Step 1: Find the median class. The total number of representatives = 130
So the middle item is 130/2 = 65
The 65th item is in the 35 – 40 class
Step 2: We can now use the formula
L = 35 Frequency before Median class = 2+14+29 = 45
Frequency of Median class = 43
Class width= 5
⎡ 65 − 45 ⎤
Median = 35 + ⎢
5 = 37.33years
⎣ 43 ⎥⎦
50% of the representatives are younger than 37.33 years and 50% are older.
The Upper and Lower Quartiles
The quartiles divide the data into quarters. 25% of the data lies below the lower
quartile and 75% above. With the upper quartile, 25% lies above it and 75% below.
25%
Lower Quartile
25%
Median
Upper Quartile
To find the Quartiles:
Find the class in which the quarter value lies
For the lower quartile use the formula:
⎡N
⎢ − Cumulative Frequency before Quarter Class
L + ⎢⎢ 4
Frequency of Quarter class
⎢
⎢⎣
⎤
⎥
⎥ Class width
⎥
⎥
⎥⎦
For the Upper use
⎡ 3N
− Cumulative Frequency before Quarter Class
⎢
L + ⎢⎢ 4
Frequency of Quarter class
⎢
⎢⎣
⎤
⎥
⎥ Class width
⎥
⎥
⎥⎦
The quartiles can be found by drawing the cumulative frequency curve (ogive)
The Inter-quartile range is the upper minus the lower quartile and is sometimes
used as a measure of dispersion.
Walter Fleming
Page 2 of 4
Measures of Central Tendency
The Mode
Mode: The most frequent value
Easy to find
Easily understood
May not be unique
Cannot be used for further analysis
Finding the Mode;
If the data is discrete (i.e. can take on whole values only) then the value that occurs
most frequently is the mode.
Example
x
f
4
2
5
5
6
21
7
17
8
9
9
2
10
1
The mode = 6 since 6 has the largest frequency of 21.
Finding the mode of a frequency distribution
1. Find the Modal class – the class with the highest frequency
2. Find D1, the difference between the largest frequency and that of the
frequency immediately before it.
3. Find D2, the difference between the largest frequency and that of the
frequency immediately after it.
4. Use the formula:
⎡ D1 ⎤
Mode = L + ⎢
⎥.C
⎣ D1 + D2 ⎦
Where
L =lower bound of modal class
C = modal class width
Example:
Age
Number of
employees
20 - 25
2
25 - 30
14
30 – 35
29
35 - 40
43
1. Modal Class is 35 – 40
2. L = 35
D1 = 43 – 29 = 14
D2 = 43 – 33 = 10
⎡ 14 ⎤
Mode = 35 + ⎢
.5 = 37.92
⎣14 + 10 ⎥⎦
Walter Fleming
Page 3 of 4
40 - 45
33
45 – 50
9
Measures of Central Tendency
The Mean
Calculated from formula
The most widely used of the measures
Used for analysis of data
Relatively easy to understand
Is influenced by extreme values
The mean is “dragged” towards
extremes
x =
∑ FX
∑F
X is the number in the top line
F is the number in the bottom line (the frequency)
Note:
Multiply F by X first and then add.
Example:
Age
Number of
employees
20 - 25
2
Walter Fleming
25 - 30
14
30 – 35
29
35 - 40
43
X
22.5
27.5
32.5
37.5
42.5
47.5
210
F
2
14
29
43
33
9
130
Mean =
4815
= 37.03
130
Page 4 of 4
FX
45
385
942.5
1612.5
1402.5
427.5
4815
40 - 45
33
45 – 50
9