Module 1
Measures of Central Tendency:
The following are the five measures of central tendency that are in common use:
i. Arithmetic mean or simple mean
ii. Median
iii. Mode
iv. Geometric mean
v. Harmonic mean
Arithmetic Mean:
Arithmetic mean of a set of observations is their sum divided by number of observations,
for example, the arithmetic mean 𝑥̅ of n observations 𝑥 , 𝑥 , 𝑥 , … , 𝑥 is given by:
𝒏
𝟏 𝟏
𝒙 = (𝒙𝟏 +𝒙𝟐 + 𝒙𝟑 + ⋯ + 𝒙𝒏 ) = 𝒙𝒊
𝒏 𝒏
𝒊 𝟏
In case the frequency distribution 𝑓 , 𝑖 = 1,2,3, … , 𝑛, where 𝑓 is the frequency
of the variable 𝑥 ,
⋯
𝑥̅ = = ∑ 𝑓 𝑥 , where 𝑁 = ∑ 𝑓
⋯
In case of grouped or continuous frequency distribution, x is taken as the as the mid value
of the corresponding class.
It may be noted that if the values of x or f are large, the calculation of mean by formula is
∑ 𝑓 𝑥 is quite time-consuming and tedious. The arithmetic is reduced to a great extent
by taking the deviations of the given values from any arbitrary point ‘A’ as explained
below:
Let 𝑑 = 𝑥 − 𝐴. Then 𝑓 𝑑 = 𝑓 (𝑥 − 𝐴) = 𝑓 𝑥 − 𝐴𝑓
Summing both sides over 𝑖 from 1 to n, we get
𝑓𝑑 = 𝑓𝑥 −𝐴 𝑓 = 𝑓 𝑥 − 𝐴𝑁
, 1 1 1 1
𝑓𝑑 = 𝑓𝑥 − 𝐴 𝑓 = 𝑓𝑥 −𝐴
𝑁 𝑁 𝑁 𝑁
1
𝑓 𝑑 = 𝑥̅ − 𝐴
𝑁
Where 𝑥̅ is the arithmetic mean of the distribution.
1
𝑥̅ = 𝐴 + 𝑓𝑑
𝑁
In case of grouped (or) continuous frequency distribution, the arithmetic is reduced to still
greater extent by taking point h is the common magnitude of class interval. In this case, we
have h𝑑 = 𝑥 − 𝐴 and proceeding exactly similarly above, we get
h
𝑥̅ = 𝐴 + 𝑓𝑑
𝑁
Problem 1:
The intelligence quotient (IQ’s) of 10 boys in a class are given below:
70, 120, 110, 101, 88, 83, 95, 98, 107, 100. Then find the mean I.Q.
Solution:
Mean I.Q of 10 boys in a class are given below:
∑
𝑋= = = = 97.2
Problem 2:
The following frequency is distribution of the number of telephone calls received in 245
successive one-minute intervals at an exchange:
No. of calls 0 1 2 3 4 5 6 7
,frequency 14 21 25 43 51 40 39 12
Obtain the mean number of calls per minute.
Solution:
No. of Calls (𝑋) Frequency (𝑓) 𝑓𝑋
0 14 0
1 21 21
2 25 50
3 43 129
4 51 204
5 40 200
6 39 234
7 12 84
𝑁 = 245 𝑓𝑋 = 922
Mean number of calls per minute at the at the exchange is given by
∑ 𝑓𝑋 922
𝑋= = = 3.763
𝑁 245
Problem 3:
Find the arithmetic mean of the following frequency distribution:
𝑋 1 2 3 4 5 6 7
𝑓 5 9 12 17 14 10 6
Solution:
, 𝑋 𝑓 𝑓𝑋
1 5 5
2 9 18
3 12 36
4 17 68
5 14 70
6 10 60
7 7 42
Total 73 299
1 299
𝑥̅ = 𝑓𝑥 = = 4.09
𝑁 73
Problem 4:
For a certain frequency table which has only been partly reproduced here, the mean was
found to be 1.46.
0 1 2 3 4 5 Total
46 ? ? 25 10 5 200
Calculate the missing frequencies.
Solution:
Let 𝑋 denote the number of accidents and let the missing frequencies corresponding to
𝑋 = 1 and 𝑋 = 2 be 𝑓 𝑎𝑛𝑑 𝑓 respectively.
No. of accidents (𝑋) Frequency (𝑓) 𝑓𝑋
0 46 0
1 𝑓 𝑓
2 𝑓 2𝑓
Measures of Central Tendency:
The following are the five measures of central tendency that are in common use:
i. Arithmetic mean or simple mean
ii. Median
iii. Mode
iv. Geometric mean
v. Harmonic mean
Arithmetic Mean:
Arithmetic mean of a set of observations is their sum divided by number of observations,
for example, the arithmetic mean 𝑥̅ of n observations 𝑥 , 𝑥 , 𝑥 , … , 𝑥 is given by:
𝒏
𝟏 𝟏
𝒙 = (𝒙𝟏 +𝒙𝟐 + 𝒙𝟑 + ⋯ + 𝒙𝒏 ) = 𝒙𝒊
𝒏 𝒏
𝒊 𝟏
In case the frequency distribution 𝑓 , 𝑖 = 1,2,3, … , 𝑛, where 𝑓 is the frequency
of the variable 𝑥 ,
⋯
𝑥̅ = = ∑ 𝑓 𝑥 , where 𝑁 = ∑ 𝑓
⋯
In case of grouped or continuous frequency distribution, x is taken as the as the mid value
of the corresponding class.
It may be noted that if the values of x or f are large, the calculation of mean by formula is
∑ 𝑓 𝑥 is quite time-consuming and tedious. The arithmetic is reduced to a great extent
by taking the deviations of the given values from any arbitrary point ‘A’ as explained
below:
Let 𝑑 = 𝑥 − 𝐴. Then 𝑓 𝑑 = 𝑓 (𝑥 − 𝐴) = 𝑓 𝑥 − 𝐴𝑓
Summing both sides over 𝑖 from 1 to n, we get
𝑓𝑑 = 𝑓𝑥 −𝐴 𝑓 = 𝑓 𝑥 − 𝐴𝑁
, 1 1 1 1
𝑓𝑑 = 𝑓𝑥 − 𝐴 𝑓 = 𝑓𝑥 −𝐴
𝑁 𝑁 𝑁 𝑁
1
𝑓 𝑑 = 𝑥̅ − 𝐴
𝑁
Where 𝑥̅ is the arithmetic mean of the distribution.
1
𝑥̅ = 𝐴 + 𝑓𝑑
𝑁
In case of grouped (or) continuous frequency distribution, the arithmetic is reduced to still
greater extent by taking point h is the common magnitude of class interval. In this case, we
have h𝑑 = 𝑥 − 𝐴 and proceeding exactly similarly above, we get
h
𝑥̅ = 𝐴 + 𝑓𝑑
𝑁
Problem 1:
The intelligence quotient (IQ’s) of 10 boys in a class are given below:
70, 120, 110, 101, 88, 83, 95, 98, 107, 100. Then find the mean I.Q.
Solution:
Mean I.Q of 10 boys in a class are given below:
∑
𝑋= = = = 97.2
Problem 2:
The following frequency is distribution of the number of telephone calls received in 245
successive one-minute intervals at an exchange:
No. of calls 0 1 2 3 4 5 6 7
,frequency 14 21 25 43 51 40 39 12
Obtain the mean number of calls per minute.
Solution:
No. of Calls (𝑋) Frequency (𝑓) 𝑓𝑋
0 14 0
1 21 21
2 25 50
3 43 129
4 51 204
5 40 200
6 39 234
7 12 84
𝑁 = 245 𝑓𝑋 = 922
Mean number of calls per minute at the at the exchange is given by
∑ 𝑓𝑋 922
𝑋= = = 3.763
𝑁 245
Problem 3:
Find the arithmetic mean of the following frequency distribution:
𝑋 1 2 3 4 5 6 7
𝑓 5 9 12 17 14 10 6
Solution:
, 𝑋 𝑓 𝑓𝑋
1 5 5
2 9 18
3 12 36
4 17 68
5 14 70
6 10 60
7 7 42
Total 73 299
1 299
𝑥̅ = 𝑓𝑥 = = 4.09
𝑁 73
Problem 4:
For a certain frequency table which has only been partly reproduced here, the mean was
found to be 1.46.
0 1 2 3 4 5 Total
46 ? ? 25 10 5 200
Calculate the missing frequencies.
Solution:
Let 𝑋 denote the number of accidents and let the missing frequencies corresponding to
𝑋 = 1 and 𝑋 = 2 be 𝑓 𝑎𝑛𝑑 𝑓 respectively.
No. of accidents (𝑋) Frequency (𝑓) 𝑓𝑋
0 46 0
1 𝑓 𝑓
2 𝑓 2𝑓
