Remark II: Assumed mean and common factor approach can equally be used to obtain the mean
2.2.3 Harmonic Mean (HM)
This is an average often used in engineering and related disciplines. It is good for calculation of average speed of machines.
For the values of variable xi, i = 1, 2, --- , n, the HM of these values is given by HM(x) =
xi
n
1 or
1 1
xi
n
Example 2.2.3.1
Using the data in example 2.2.2.1, obtain the Harmonic mean.
HM(x) =
xi
n 1
= 10
1/12 + 1/7 + 1/2 + 1/6 + 1/13 + 1/17 + 1/14 + 1/5 + 1/9 + 1/4
= 10
0.0833 + 0.1429 + 0.5 + 0.1667 + 0.0769 + 0.0588 + 0.0714 + 0.2 + 0.1111 + 0.25 = 10
1.6611
HM(x) = 6.0201
li 2.3 Mode and Median
The MODE is the value which occurs most frequently in a set of data. For a data set in which no measured values are repeated, there is no mode. Otherwise, a distribution can be Unimodal, Bimodal or Multimodal according to the data set.
From a grouped data with frequency distribution, the mode is determined by either a. graphical method through the use of Histogram; and
b. the use of formula.
a. In the graphical method, the following steps are to be taken:
i. Draw the histogram of the given distribution.
ii. Identify the highest bar (the modal class).
iii. Identify the two linking bars to the modal class, i.e the bar before and after the modal class.
iv. Use the two flanking bars to draw diagonals on the modal class.
v. Locate the point of intersection of the diagonals drawn in step (iv) above, and vertically draw a line from the point of intersection to the boundary axis which gives the desired mode.
b. By formula, the mode is defined by Mode = L1 + 1 C
1 + 2
where
L1 = Lower class boundary of the modal class
1 = Modal class frequency – Frequency of the class before the modal class.
2 = Modal class frequency – Frequency of the class after the modal class.
C = Modal class size
The MEDIAN of a data set is the value of the middle item of the data when all the items in the data set are arranged in an array form (either ascending or descending order).
From an ungrouped data, the position (Median) is located by (N + 1)/2, where N is the total number of items in the data set.
In a grouped data, the position of the Median is located by N/2. Here, the specific value of the Median is determined by either
( )
lii
a. graphical method through the use of Ogive (Cumulative frequency curve)or b. the use of formula
a. In the graphical method, the following procedure is used:
i. Draw the Ogive of the given distribution.
ii. Locate the point N/2 on the cumulative frequency axis of the Ogive.
iii. Draw a parallel line through the value of N/2 to the curve and then draw a perpendicular line from curve intercept to the boundary axis in order to get the Median value.
b. The Median is obtained by the use of the following formula:
Median = L1 + N/2 – (Fi) C fi
where L1 = Lower class boundary of median class.
N = Total number of items in the data set.
Fi = Summation of all frequencies before the median class.
fi = Frequency of the median class.
C = Median class size or interval.
Worked Examples:
Example 2.3.1: Case I (n is odd and discrete variable is involved)
The following are the data generated from the sales of oranges within eleven days: 4, 7, 12, 3, 5, 3, 6, 2, 3, 1, 3, calculate both the mode and median.
Solution
Mode = Most occurring frequency
Mode = 3
Median = {n + 1}th position 2
i.e 1, 2, 3, 3, 3, 3, 4, 5, 6, 7, 12
Median (n = 11) = 11 + 1 = 12 = 6th position
2 2
The median = 3
Example 2.3.2: (n is odd and continuous variable is involved)
The following are the data generated from the measurement of an iron-rod; 2.5, 2.0, 2.1, 3.5, 2.5,
( )
liii 2.5, 1.0, 2.5, 2.2, calculate both the mode and median.
Solution
1.0, 2.0, 2.1, 2.2, 2.5, 2.5, 2.5, 2.5, 3.5 Mode = 2.5
Median (n = 9) = n + 1 = 9 + 1 = 10 = 5th position
2 2 2
The median = 2.5
Example 2.3.3: (n is even and discrete variable is involved)
The following are the data generated from the sales of orange within 12 days 4, 7, 12, 3, 5, 3, 6, 2, 3, 1, 3, 4, calculate the mode and median.
Solution
Mode = Most occurrence frequency
Mode = 3
Median= (n + 1)th position 2
If n = 12, 12 + 1 = 13 = 6.5th position
2 2
Rearrange the data first and take the middle value.
1, 2, 3, 3, 3, 3, 4, 4, 5, 6, 7, 12
The 6.5th position gives 3 + 4 = 3.5 2
Example 2.3.4 (n is even and continuous variable is involved):
The following are the data generated from the measurement of an iron-rod; 2.5, 2.0, 2.1, 3.5, 2.5, 2.5, 1.0, 2.5, 2.2, 2.6, calculate the mode and median
Solution
Mode = Most occurring frequency Mode = 2.5
Median= (n + 1)th position 2
If n = 12, 12 + 1 = 13 = 6.5th position
2 2
Rearrange the data first and take the middle value.
1.0, 2.0, 2.1, 2.2, 2.5, 2.5, 2.5, 2.5, 2.6, 3.5
liv
The 6.5th position gives 2.5 + 2.5 = 2.5 2
Median = 2.5
Example 2.3.5: Case III: Grouped data by graphical method Use the graphical method to determine the mode of the following data
Tyre range Frequency C. F Class Boundaries
2 – 4 5 – 7 8 – 10 11 – 13 14 – 16
3 4 6 2 1
3 7 13 15 20
1.5 – 4.5 4.5 – 7.5 7.5 – 10.5 10.5 – 13.5 13.5 – 16.5 20
Solution
- Draw the histogram of the given table
- Interpolate on the modal class as discussed earlier in order to obtain the mode.
Mode = 8.5
Example 2.3.6: Case IV: Graphical method
Use graphical method to find the median using case III data Solution:
4.5 7.5 10.5 13.5 16.5
1.5 0 1
2 3 4 5 6 7
Class Boundaries
FREQUENCY
lv
Tyre range Frequency C. F Less than Class
Boundary 2 – 4
5 – 7 8 – 10 11 – 13 14 – 16
3 4 6 2 1
3 7 13 15 20
4.5 7.5 10.5 13.5 16.5 20
Median frequency = N/2 = 20/2 = 10
Median = 9.0
Example 2.3.7: Graphical method to find mode and median.
Determine the mode and median by graphical method from the following data.
Class interval Frequency(f) C. F
2 – 4 4 – 6 6 – 8 8 – 10 10 – 12
3 4 6 7 2
3 7 13 20 22 22
lvi Solution
Mode = 8.50
To find the Median, we proceed as follows:
Table Frequency (f) C. F Less than Boundary
2 – 4 4 – 6 6 – 8 8 – 10 10 – 12
3 4 6 7 2
3 7 13 20 22
4 6 8 10 12 22
4 6 8 10 12
2 0 1
2 3 4 5 6 7
Class Boundaries
FREQUENCY
0 5 10
15 20 25
4 6 8 10 12
2
Cummulative Frequency
lvii
Median frequency = N/2 = 22/2 = 11
Median = 7.8
Example 2.3.8: Case III: Grouped data using formula
The following are the data generated on the sales of electric data Class interval Frequency
2 – 4 3
4 – 6 4
6 – 8 6
8 – 10 7
10 – 12 2
Determine the mode and median using the formula.
Solution
Class Frequency Cumulative Frequency
2 – 4 4 – 6 6 – 8 8 – 10 10 – 12
3 4 6 7 2
3 7
13 Median class 20 Modal class 22
22
N = 22; N/2 = 22/2 = 11
Mode = L1 + 1 C
1 + 2
where
L1 = Lower class boundary of the modal class
1 = Modal class frequency – Frequency of the class before the modal class.
2 = Modal class frequency – Frequency of the class after the modal class.
C = Modal class size
L1 = 8, 1 = 7 – 6 = 1, 2 = 7 – 2 = 5, C = 2 Mode = 8 + 1 (2)
( )
( )
lviii 1 + 5
= 8 + 1 (2) 6
= 8 + 1/3
= 8 + 0.33 = 8.33
Mode = 8.33
Median = L1 + N/2 – (Fi) C fi
where L1 = Lower class boundary of the median class fi = Frequency of the median class
C = Median class size
Fi = Summation of all frequencies before the median class L1 = 6, Fmed = 6, C = 2, fi = 3 + 4 = 7, N = 22
Median = 6 + 22/2 – 7 (2) 6
Median = 6 + 11 – 7 (2)
6
Median = 6 + 4 (2)
6
Median = 6 + 4
3
Median = 6 + 1.3
Median = 7.3
Example 2.3.9
The following are the data generated on the sales of tyres:
Tyre range 2 – 4 4 – 6 6 – 8 8 – 10 10 – 12
Frequency 3 4 6 2 5
Calculate the mode and median.
Solution
( )
( )
( ) ( )
( )
lix
Tyre range Frequency Cumulative Frequency
2 – 4 4 – 6 6 – 8 8 – 10 10 – 12
3 4 6 2 5
3 7
13 Median/Modal class 15
20 20
N = 20; N/2 = 20/2 = 10
Mode = L1 + 1 C
1 + 2
L1 = 6.0, 1 = 6 – 4 = 2, 2 = 6 – 2 = 4, C = 3 (i.e 4.5 – 1.5) Mode = 6.0 + 2 (3)
2 + 4
= 6.0 + 1 Mode = 7.0
Median = L1 + N/2 – (Fi) C fi
L1 = 6.0, Fmed = 6, C = 3, Fi = fi = 3 + 4 = 7, N = 20 Median = 6.0 + 20/2 – 7 (3)
6
Median = 6.0 + 10 – 7 (3)
6
Median = 6.0 + 3 (3)
6
Median = 6.0 + 3
2
Median = 6.0 + 1.5
( )
( )
( )
( )
( )
( )
lx
Median = 7.5