Unit II
Measures of Central tendency
Measures of Central Tendency
A measure of central tendency is a statistical summary that represents the center point of the dataset. It indicates where most values in a distribution fall. It is also called as measure of central location.
The three most common measure of central tendency are Mean, Median, and Mode.
Definition
According to Prof Bowley, “Measures of central tendency (averages) are statistical constants which enable us to comprehend in a single effort the significance of the whole.”
Requisites of a good measure of central tendency
 It should be rigidly defined
 It should be simple to understand and easy to calculate
 It should be based upon all values of given data.
 It should be capable of further mathematical treatment.
 It should have sampling stability.
 It should be not be unduly affected by extreme values.
 The mean is the arithmetic average, also called as arithmetic mean.
 Mean is very simple to calculate and is most commonly used measure of the center of data.
 Means is calculated by adding up all the values and divided by the number of observation.
Computation of sample mean 
If X1, X2, ………………Xn are data values then arithmetic mean is given by
Computation of the mean for ungrouped data
Example 1 – The marks obtained in 10 class test are 25, 10, 15, 30, 35
The mean = X = 25+10+15+30+35 = 115 =23
5 5
Analysis – The average performance of 5 students is 23. The implication is that students who got below 23 did not perform well. The students who got above 23 performed well in exam.
Example 2 – Find the mean
Xi  9  10  11  12  13  14  15 
Freq (Fi)  2  5  12  17  14  6  3 
Xi  Freq (Fi)  XiFi 
9  2  18 
10  5  50 
11  12  132 
12  17  204 
13  14  182 
14  6  84 
15  3  45 
 Fi = 59  XiFi= 715 



Then, N = ∑ fi = 59, and ∑fi Xi=715
X = 715/59 = 12.11
Mean for grouped data/ Weighted Arithmetic Mean
Grouped data are the data that are arranged in a frequency distribution
Frequency distribution is the arrangement of scores according to category of classes including the frequency.
Frequency is the number of observations falling in a category
The formula in solving the mean for grouped data is called midpoint method. The formula is
Where, X = Mean
Xm = midpoint of each class or category
f = frequency in each class or category
∑f Xm = summation of the product of fXm
Example 3 – The following data represent the income distribution of 100 families. Calculate mean income of 100 families?
Income  3040  4050  5060  6070  7080  8090  90100 
No. Of families  8  12  25  22  16  11  6 
Solution:
Income  No. Of families  Xm (Mid point)  FXm 
3040  8  35  280 
4050  12  34  408 
5060  25  55  1375 
6070  22  65  1430 
7080  16  75  1200 
8090  11  85  935 
90100  6  95  570 
 n = 100 
 ∑f Xm = 6198 
X = ∑f Xm/n = 6330/100 = 63.30
Mean = 63.30
Example 4 – Calculate the mean number of hours per week spent by each student in texting message.
Time per week  0  5  5  10  10  15  15  20  20  25  25 – 30 
No. Of students  8  11  15  12  9  5 
Solution:
Time per week (X)  No. Of students (F)  Mid point X  XF 
0  5  8  2.5  20 
5 – 10  11  7.5  82.5 
10  15  15  12.5  187.5 
15  20  12  17.5  210 
20  25  9  22.5  202.5 
25 – 30  5  27.5  137.5 
 60 
 840 
Mean = 840/60 = 14
Example 5 –
The following table of grouped data represents the weights (in pounds) of all 100 babies born at a local hospital last year.
Weight (pounds)  Number of Babies 
[3−5)  8 
[5−7)  25 
[7−9)  45 
[9−11)  18 
[11−13)  4 
Solution:
Weight (pounds)  Number of Babies  Mid point X  XF 
[3−5)  8  4  32 
[5−7)  25  6  150 
[7−9)  45  8  360 
[9−11)  18  10  180 
[11−13)  4  12  48 
 100 
 770 
Mean = 770/100 = 7.7
Merits of mean
 It is rigidly defined
 It is easy to understand and easy to calculate
 It is based upon all values of the given data
 It is capable of future mathematical treatment
 It is not much affected by sampling fluctuation
Demerits of mean
 It cannot be calculated if any observation are missing
 It cannot be calculated for open end classes
 It is effected by extreme values
 It cannot be located graphically
 It may be number which is not present in the data
 The points or value that divides the data into two equal parts
 Firstly, the data are arranged in ascending or descending order .
 The median is the middle number depending on the data size.
 When the data size is odd, the median is the middle value
 When the data size is even, median is the average of the middle two values
 It is also known as middle score or 50th percentile
For ungrouped data median is calculated by (n+1)th value
2
Example 1 – find the median score of 7 students in science class
Score = 19, 17, 16, 15, 12, 11, 10
Median = (7+1)/2 = 4th value
Median = 15
Find the median score of 8 students in science class
Score = 19, 17, 16, 15, 12, 11, 10, 9
Median = (8+1)/2 = 4.5th value
Median = (15+12)/2 = 13.5
Example 2 – Find the median of the table given below
Marks obtained  No. Of students 
20  6 
25  20 
28  24 
29  28 
33  15 
38  4 
42  2 
43  1 
Solution:
Marks obtained  No. Of students  Cf 
20  6  6 
25  20  26 (20+6) 
28  24  50 (26+24) 
29  28  78 
33  15  93 
38  4  97 
42  2  99 
43  1  100 
Median = (n+1)/2 = 100+1/2 = 50.5
Median = (28+29)/2 = 28.5
Median of grouped data
Formula
MC = median class is a category containing the n/2
Lb = lower boundary of the median class
Cfp = cumulative frequency before the median class if the scores are arranged from lowest to highest value
Fm = frequency of the median class
c.i = size of the class interval
Ex calculate the median
Example 3
Calculate the median
Marks  No. Of students 
04  2 
59  8 
1014  14 
1519  17 
2024  9 
Solution:
Marks  No. Of students  CF 
04  2  2 
59  8  10 
1014  14  24 
1519  17  41 
2024  9  50 
 50 

n = 50
n = 50/2= 25
2
The category containing n/2 is 15 19
Lb = 15
Cfp = 24
f = 17
Ci = 4
Median = 15 + 2524 *4 = 15.23
17
Example 4  Given the below frequency table calculate median
X  60 – 70  70 – 80  80 90  90100 
F  4  5  6  7 
Solution:
X  F  CF 
60  70  4  4 
70  80  5  9 
80  90  6  15 
90  100  7  22 
n = 22
n = 22/2= 11
2
The category containing n+1/2 is 80  90
Lb = 80
Cfp = 9
f = 6
Ci = 10
Median = 80 + 119 *10 = 83.33
6
Example 5– Calculate the median of grouped data
Class interval  13  35  57  79  911  1113 
Frequency  4  12  13  19  7  5 
Solution:
CI  F  CF 
13  4  4 
35  12  16 
57  13  29 
79  19  48 
911  7  55 
1113  5  60 
n = 60
n = 60/2= 30
2
The category containing n+1/2 is 79
Lb = 7
Cfp = 29
f = 19
Ci = 2
Median = 7 + 3029 *2 = 7.105
19
Merits of median
 It is rigidly defined
 It is easy to understand and easy to calculate
 It is not effected by extreme values
 It is not much affected by sampling fluctuation
 It can be located graphically
Demerits of median
 It is not based upon all values of the given data
 It is difficult to calculate increasing order data size
 It is not capable of further mathematical treatment.
The mode is denoted Mo, is the value which occurs most frequently in a set of values. Croxton and Cowden defined it as “the mode of a distribution is the value at the point armed with the item tends to most heavily concentrated. It may be regarded as the most typical of a series of value”
Mode for ungrouped data
Example 1 Find the mode of scores of section A
Scores = 25, 24, 24, 20, 17, 18, 10, 18, 9, 7
Solution – Mode is 24, 18 as both have occurred twice.
Mode for grouped data
Mode = L1 + (L2 – L1) d1
d1 +d2
L1= lower limit of the modal class,
L2= upper limit of the modal class‟
d1 =fmf0 and d2=fmf1
Where fm= frequency of the modal class,
f0 = frequency of the class preceding to the modal class,
f1= frequency of the class succeeding to the modal class.
Example 2 – Find the mode
Seconds  Frequency 
51  55  2 
56  60  7 
61  65  8 
66  70  4 
The group with the highest frequency is the modal group:  6165
D1 = 87 = 1
D2 = 84 = 4
Mode = L1 + (L2 – L1) d1
d1 +d2
mode = 61 + (6561) 1 = 61+4 (1/5) = 61.8
1+4
Mode = 61.8
Example 3  In a class of 30 students marks obtained by students in science out of 50 is tabulated below. Calculate the mode of the given data.
Marks obtained  No. Of students 
10 20  5 
20 – 30  12 
30 – 40  8 
40  50  5 
Solution:
The group with the highest frequency is the modal group:  20 30
D1 = 12  5 = 7
D2 = 12  8 = 4
Mode = L1 + (L2 – L1) d1
d1 +d2
mode = 20 + (3020) 7 = 20+10 (7/11) = 26.36
7+4
Mode = 61.8
Example 4 Based on the group data below, find the mode
Time to travel to work  Frequency 
1 – 10  8 
11 20  14 
21 – 30  12 
31 – 40  9 
41  50  7 
Solution:
The group with the highest frequency is the modal group:  11  20
D1 = 14  8 = 6
D2 = 14  12 = 2
Mode = L1 + (L2 – L1) d1
d1 +d2
mode = 11 + (2011) 6 = 11+9 (6/8) = 17.75
6+2
Example 5 –
Compute the mode from the following frequency distribution
CI  F 
7071  2 
6869  2 
6667  3 
6465  4 
6263  6 
6061  7 
5859  5 
Solution:
The group with the highest frequency is the modal group:  60  61
D1 = 7  6 = 1
D2 = 7  5 = 2
Mode = L1 + (L2 – L1) d1
d1 +d2
mode = 60 + (6160) 1 = 60+1 (1/3) 60.85
1+2
Merits of mode
 It is easy to understand & easy to calculate
 It is not affected by extreme values or sampling fluctuations.
 Even if extreme values are not known mode can be calculated.
 It can be located just by inspection in many cases.
 It is always present within the data.
Demerits of mode
 It is not rigidly defined.
 It is not based upon all values of the given data.
 It is not capable of further mathematical treatment.
Geometric mean
Geometric mean is a type of mean or average, which indicates the central tendency of a set of numbers by using the product of their values.
Definition
The Geometric Mean (G.M) of a series containing n observations is the nth root of the product of the values.
For ungrouped data
Geometric Mean, GM = Antilog ∑logxi
N
Example 1 – find the G.M of the values
X  Log X 
45  1.653 
60  1.778 
48  1.681 
65  1.813 
Total  6.925 
GM = Antilog ∑logxi
N
= Antilog 6.925/4
= Antilog 1.73
= 53.82
For grouped data
Geometric Mean, GM = Antilog ∑ f logxi
N
Example 2 – calculate the geometric mean
X  f 
60 – 80  22 
80 – 100  38 
100 – 120  45 
120 – 140  35 


Solution
X  f  Mid X  Log X  f log X 
60 – 80  22  70  1.845  40.59 
80 – 100  38  90  1.954  74.25 
100 – 120  45  110  2.041  91.85 
120 – 140  35  130  2.114  73.99 
Total  140 

 280.68 
GM = Antilog ∑ f logxi
N
= antilog 280.68/140
= antilog 2.00
GM = 100
Example 3 – calculate geometric mean
Class  Frequency 
24  3 
46  4 
68  2 
810  1 
Solution
Class  Frequency  x  Log x  Flogx 
24  3  3  1.0986  3.2958 
46  4  5  1.2875  6.4378 
68  2  7  0.5559  3.8918 
810  1  9  0.2441  2.1972 
 10 

 15.8226 
GM = Antilog ∑ f logxi
N
= antilog 15.8226/10
= antilog 1.5823
GM = 4.866
Harmonic mean
Harmonic mean is quotient of “number of the given values” and “sum of the reciprocals of the given values
For ungrouped data
Example 1  Calculate the harmonic mean of the numbers 13.2, 14.2, 14.8, 15.2 and 16.1
Solution
X  1/X 
13.2  0.0758 
14.2  0.0704 
14.8  0.0676 
15.2  0.0658 
16.1  0.0621 
Total  0.3147 
H.M of X = 5/0.3147 = 15.88
Example 2  Find the harmonic mean of the following data {8, 9, 6, 11, 10, 5} ?
X  1/X 
8  0.125 
9  0.111 
6  0.167 
11  0.091 
10  0.100 
5  0.200 
Total  0.794 
H.M of X = 6/0.794 = 7.560
For grouped data
Harmonic Mean ∑X =
∑∑
∑∑
Example 3  Calculate the harmonic mean for the below data
Marks  3039  4049  5059  6069  7079  8089  9099 
F  2  3  11  20  32  25  7 
Solution
Marks  X  F  F/X 
3039  34.5  2  0.0580 
4049  44.5  3  0.0674 
5059  54.4  11  0.2018 
6069  64.5  20  0.3101 
7079  74.5  32  0.4295 
8089  84.5  25  0.2959 
9099  94.5  7  0.0741 
Total 
 100  1.4368 
HM = 100/1.4368 = 69.59
Example 4 – find the harmonic mean of the given class
Ages  4  5  6  7 
No. Of students  6  4  10  9 
Solution
X  F  f/x 
4  6  1.50 
5  4  0.80 
6  10  1.67 
7  9  1.29 
 29.00  5.25 
HM = 29/5.25 = 5.5
Example 5 – calculate harmonic mean
Class  Frequency 
24  3 
46  4 
68  2 
810  1 
Solution
Class  Frequency  x  f/x 
24  3  3  1 
46  4  5  0.8 
68  2  7  0.28 
810  1  9  0.11 
 10 
 2.19 
Harmonic mean = 10/2.19 = 4.55
There are three quartiles, i.e. Q1, Q2 and Q3 which divide the total data into four equal parts when it has been orderly arranged. Q1, Q2 and Q3 are termed as first quartile, second quartile and third quartile or lower quartile, middle quartile and upper quartile, respectively. The first quartile, Q1, separates the first onefourth of the data from the upper three fourths and is equal to the 25th percentile. The second quartile, Q2, divides the data into two equal parts (like median) and is equal to the 50th percentile. The third quartile, Q3, separates the first threequarters of the data from the last quarter and is equal to 75th percentile.
Calculation of Quartiles:
The calculation of quartiles is done exactly in the same manner as it is in case of the calculation of median.
The different quartiles can be found using the formula given below:
Qi = l1 + i= 1,2,3
Where,
L1 = lower limit of ith quartile class
L2 = upper limit of ith quartile class
c = cumulative frequency of the class preceding the ith quartile class
f = frequency of ith quartile class.
Example 1 : Calculate Q1, Q2 and Q3 from the following data given below:
Day  Frequency 
1  20 
2  35 
3  25 
4  12 
5  10 
6  23 
7  18 
8  14 
9  30 
10  40 
Solution:
Arrange the frequency data in ascending order
Day  Frequency 
1  10 
2  12 
3  14 
4  18 
5  20 
6  23 
7  25 
8  30 
9  35 
10  40 
First quartile (Q1)
Qi= [i * (n + 1) /4] th observation
Q1= [1 * (10 + 1) /4] th observation
Q1 = 2.75 th observation
Thus, 2.75 th observation lies between the 2nd and 3rd value in the ordered group, between frequency 12 and 14
First quartile (Q1) is calculated as
Q1 = 2nd observation +0.75 * (3rd observation  2nd observation)
Q1 = 12 + 0.75 * (14 – 12) = 13.50
Third quartile (Q3)
Qi= [i * (n + 1) /4] th observation
Q3= [3 * (10 + 1) /4] th observation
Q3 = 8.25 th observation
So, 8.25 th observation lies between the 8th and 9th value in the ordered group, between frequency 30 and 35
Third quartile (Q3) is calculated as
Q3 = 8th observation +0.25 * (9th observation – 8th observation)
Q3 = 30 + 0.25 * (35 – 30) = 31.25
Example 3
Age in years  40 44  45 – 49  50 – 54  55  59  60 – 64  65  69 
Employees  5  8  11  10  9  7 
Solutions:
In the case of Frequency Distribution, Quartiles can be calculated by using the formula:
Class interval  F  Class boundaries  CF 
40 44  5  39.5 – 44.5  5 
45 – 49  8  44.5 – 49.5  13 
50 – 54  11  49.5 – 54.5  24 
55 – 59  10  54.5 – 59.5  34 
60 – 64  9  59.5 – 64.5  43 
65 – 69  7  64.5 – 69.5  50 
Total  50 


First quartile (Q1)
Qi= [i * (n ) /4] th observation
Q1 = [1*(50)/4]th observation
Q1 = 12.50th observation
So, 12.50th value is in the interval 44.5 – 49.5
Group of Q1 = 44.5 – 49.5
Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3
Q1 = (44.5 + ( 5/ 8)* (1* (50/4) – 5)
Q1 = 49.19
Third quartile (Q3)
Qi= [i * (n) /4] th observation
Q3= [3 * (50) /4] th observation
Q3 = 37.5th observation
So, 37.5th value is in the interval 59.5 – 64.5
Group of Q3 = 59.5 – 64.5
Qi = (I + (h / f) * ( i * (N/4) – c) ; i = 1,2,3
Q3 = (59.5 + ( 5/ 9)* (3* (50/4) – 34)
Q3 = 61.44
Deciles are the partition values which divide the arranged data into ten equal parts. There are nine deciles i.e. D1, D2, D3……. D9 and 5th decile is same as median or Q2, because it divides the data in two equal parts.
Calculation of Deciles:
The calculation of deciles is done exactly in the same manner as it is in case of calculation of median.
The different deciles can be found using the formula given below:
Di = l1 + i= 1,2,3….9
Where,
l1 = lower limit of ith quartile class
l2 = upper limit of ith quartile class
c = cumulative frequency of the class preceding the ith quartile class
f = frequency of ith quartile class.
Example 1: Calculate Q1, D7 and P20 from the following data: 3, 13, 11, 11, 5, 4, 2
Solution: Arranging observations in the ascending order we get
2, 3, 4, 5, 11, 11, 13
Here, n = 7
Q1 = ()th value of the observation
= ()th Value of the observation
= 2nd Value of the observation
= 3
D3 = ()th value of the observation
= ()th value of the observation
= (2.4)th Value of the observation
= 2nd observation + 0.4 (3rd – 2nd)
= 3 + 0.4(4 – 3)
= 3 + 0.4
= 3.4
P20 = ()th value of the observation
= ()th value of the observation
= (1.6)th value of the observation
= 1st observation + 0.6 (2nd – 1st )
= 2 + 0.6(3 – 2)
= 2 + 0.6
= 2.6
Example 2: Calculate D7 from the following data:
Class  2  4  4  6  6  8  8  10 
Frequency  3  4  2  1 
Solutions:
In the case of Frequency Distribution, Deciles can be calculated by using the formula:
Di = l1 +
Class interval  F  CF 
2  4  3  3 
4  6  4  7 
6  8  2  9 
8  10  1  10 
Total  n = 10 

Here n = 10
Class with th value of the observation in CF column
= th value of the observation in CF column
= 7th value of the observation in CF column and it lies in the class 6 – 8
Therefore, D7 class is 6 – 8
The lower boundary point of 6 – 8 is 6.
Therefore, L = 6
D7 = L +
= 6 + x 2
= 6 + 0
= 6
Percentiles are the values which divide the arranged data into hundred equal parts. There are 99 percentiles i.e. P1, P2, P3, ……. P99.
The 50th percentile divides the series into two equal parts and P50 = D5 = Median.
Similarly, the value of Q1 = P25 and value of Q3 = P75
Calculation of Percentiles:
The different percentiles can be found using the formula given below:
pi = l1 + i= 1,2,3…………….99
Where,
l1 = lower limit of ith quartile class
l2 = upper limit of ith quartile class
c = cumulative frequency of the class preceding the ith quartile class
f = frequency of ith quartile class.
Example 2: Calculate P20 from the following data:
Class  2  4  4  6  6  8  8  10 
Frequency  3  4  2  1 
Solutions:
In the case of Frequency Distribution, Percentiles can be calculated by using the formula:
pi = l1 +
Class interval  F  CF 
2  4  3  3 
4  6  4  7 
6  8  2  9 
8  10  1  10 
Total  n = 10 

Here n = 10
Class with th value of the observation in CF column
= th value of the observation in CF column
= 2th value of the observation in CF column and it lies in the class 2  4
Therefore, P20 class is 2 – 4
The lower boundary point of 2 – 4 is 2.
Therefore, L = 2
P20 = L +
= 2 + x 2
= 2 + 1.3333
= 3.3333
References
 B.N Gupta – Statistics
 S.P Singh – statistics
 Gupta and Kapoor – Statistics
 Yule and Kendall – Statistics method