Table of Contents
Arithmetic Mean:
Here we will learn all the Arithmetic Mean Formula With Example. Arithmetic mean is defined as the sum of all values divided by total number of values. Arithmetic mean is also called arithmetic average. It is most commonly used measure of central tendency.
Arithmetic averages are of two types:
Calculation of Simple Arithmetic Mean/Average
We calculate simple arithmetic mean for the Individual Series, Discrete Series and Continuous Series. Following are the methods for finding arithmetic mean:
Direct Method
Individual Series
Let X is the variable which takes values x_{1}, x_{2}, x_{3}, …, x_{n} over ‘n’ times, then arithmetic mean, simply the mean of X, denoted by bar over the variable X is given by,
\[\overline{X}=\frac{{{x}_{1}}+{{x}_{2}}+{{x}_{3}}+…+{{x}_{n}}}{n}=\frac{\sum\limits_{i=1}^{n}{{{x}_{i}}}}{n}\]
Discrete Series
\[\overline{X}=\frac{\sum{fx}}{N=\sum{f}}\]
∑fx = Sum of the product of the values and their corresponding frequencies
N = Sum of the frequencies i.e., ∑f or total number of observations.
Continuous Series
\[\overline{X}=\frac{\sum{fm}}{N=\sum{f}}\]
∑fm = Sum of the product of mid values and their corresponding frequencies
N = Sum of the frequencies i.e., ∑f or total number of observations.
Example 01 |
Calculate the mean of following data. Marks obtained by 6 students given: 20, 15, 23, 22, 25, 20.
Solution:
Mean marks,
\[\overline{X}=\frac{{{x}_{1}}+{{x}_{2}}+…+{{x}_{n}}}{n}\]
\[=\frac{20+15+23+22+25+20}{6}\]
\[=\frac{125}{6}=20.83\]
Example 02 |
Six month income of a departmental store is given below. Find mean income of store.
Month | Jan | Feb | Mar | Apr | May | June |
Income($) | 25000 | 30000 | 45000 | 20000 | 25000 | 20000 |
Solution:
n = Total number of items (observations) = 6
Total income = ∑ x_{i} = $(25000 + 30000 + 45000 + 20000 + 25000 + 20000)
= $165000
Mean Income,
\[=\frac{\sum{{{x}_{i}}}}{n}=\frac{$ 165000}{6}=$ 27500\]
Example 03 |
Calculate the arithmetic mean by direct method from the following data
Wage | 10 | 20 | 30 | 40 | 50 |
No. of Workers | 4 | 5 | 3 | 2 | 5 |
Solution:
Let us denote the wages by x and number of workers by f.
Wages (x) | No. of Workers (f) | fx |
10 | 4 | 40 |
20 | 5 | 100 |
30 | 3 | 90 |
40 | 2 | 80 |
50 | 5 | 250 |
∑ f = 19 | ∑ fx = 560 |
Hence, mean wage of the workers,
\[\overline{X}=\frac{\sum{fx}}{\sum{f}}=\frac{560}{19}=29.47\]
Example 04 |
Calculate the missing value when its mean is 115.86.
Wage | 110 | 112 | 113 | 117 | – | 125 | 128 | 130 |
No. of Workers | 25 | 17 | 13 | 15 | 14 | 8 | 6 | 2 |
Solution:
Let us denote the wages by x, number of workers by f and missing item by ‘a’.
Wages (x) | No. of Workers (f) | fx |
110 | 25 | 2750 |
112 | 17 | 1904 |
113 | 13 | 1469 |
117 | 15 | 1755 |
a | 14 | 14a |
125 | 8 | 1000 |
128 | 6 | 768 |
130 | 2 | 260 |
∑ f = 100 | ∑ fx = 9906 + 14a |
\[\overline{X}=\frac{\sum{fx}}{\sum{f}}\]
\[\Rightarrow 115.86=\frac{9906+14a}{100}\]
\[\Rightarrow 115.86\times 100=9906+14a\]
\[\Rightarrow 14a=11586-9906\]
\[\Rightarrow 14a=1680\]
\[\Rightarrow a=\frac{1680}{14}\]
\[\therefore a=120\]
Example 05 |
Find the mean for the following distribution by using direct method.
Class Interval | 84-90 | 90-96 | 96-102 | 102-108 | 108-114 |
frequency | 8 | 12 | 15 | 10 | 5 |
This is the case of continuous series. Let ‘m’ be the mid-value and ‘f’ be the frequency.
Class interval | Mid-value (m) | Frequency (f) | fx |
84-90 | 97 | 8 | 776 |
90-96 | 93 | 12 | 1116 |
96-102 | 99 | 15 | 1485 |
102-108 | 105 | 10 | 1050 |
108-114 | 111 | 5 | 555 |
∑ f = 50 | ∑ fx = 4982 |
Hence, the mean,
\[\overline{X}=\frac{\sum{fm}}{\sum{f}}=\frac{4982}{50}=99.64\]
Short Cut Method
Individual Series
The arithmetic mean can also be calculated by taking deviations from any arbitrary points. Steps of this method are given below.
Step 1: Assume any one value as a mean which is called arbitrary average (A).
Step 2: Find the difference (deviations) of each value from arbitrary average.
d = x_{i} – A
Step 3: Add all differences to get ∑ d.
Step 4: Use following equation and compute the mean value.
\[\overline{X}=A+\frac{\sum{d}}{n}\]
Discrete Series
\[\overline{X}=A+\frac{\sum{fd}}{\sum{f}}\]
Continuous Series
\[\overline{X}=A+\frac{\sum{fd}}{\sum{f}}\]
Where,
n = Total number of observations.
∑ d = Total deviation value
d = Deviation of item from the assumed mean.
A = Assumed mean.
∑ fd = Sum of products of deviations and their corresponding frequencies.
Example 01 |
Determine the average salary of a staff from the following data relating to the monthly salaries of the teaching staff of a college by using short cut method.
Salary ($) | 2200 | 2500 | 3000 | 3700 | 4500 |
No. of Staff | 5 | 10 | 15 | 7 | 3 |
Solution:
Let assumed mean A = 3000
x | f | D = (x – A) | fd |
2200 | 5 | -800 | -4000 |
2500 | 10 | -500 | -5000 |
3000 | 15 | 000 | 000 |
3700 | 7 | 700 | 4900 |
4500 | 3 | 1500 | 4500 |
∑ f = 40 | ∑ fd = 400 |
We have,
\[\overline{X}=A+\frac{\sum{fd}}{\sum{f}}=3000+\frac{400}{40}=3000+10=3010\]
Example 02 |
Calculate the average marks obtained by BCA students in Mathematics paper by short cut method
Class of Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
No. of students | 5 | 3 | 7 | 25 | 20 |
Solution:
Let A = 25
Marks of Students | No. of Students(f) | Mid-values(m) | d = m – 25 | fd |
0-10 | 5 | 5 | -20 | -100 |
10-20 | 3 | 15 | -10 | -30 |
20-30 | 7 | 25 | 0 | 00 |
30-40 | 25 | 35 | 10 | 250 |
40-50 | 20 | 45 | 20 | 400 |
∑ f = 60 | ∑ fd = 520 |
\[\overline{X}=A+\frac{\sum{fd}}{\sum{f}}\]
\[=25+\frac{520}{60}=25+8.67=33.67\]
Step Deviation Method
This method is extension of short cut method. This method is used when the figures of deviations appear to be big and divisible by a common factor.
Individual Series
\[\overline{X}=A+\frac{\sum{{{d}^{‘}}}}{N}\times c\]
Discrete Series
\[\overline{X}=A+\frac{\sum{f{{d}^{‘}}}}{\sum{f}}\times c\]
Continuous Series
\[\overline{X}=A+\frac{\sum{f{{d}^{‘}}}}{\sum{f}}\times c\]
Where,
c = common factor by which each of the deviation is divided.
d’ = the deviation from the assumed average divided by the common factor.
Example 01 |
Calculate the arithmetic mean by means of step deviation method
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
No. of students | 20 | 24 | 40 | 36 | 20 |
Solution:
Let Assumed average A = 25
Marks | Mid-values(m) | f | A = 25, d = m – 25 | d’ = (d/10) | fd’ |
0-10 | 5 | 20 | -20 | -2 | -40 |
10-20 | 15 | 24 | -10 | -1 | -24 |
20-30 | 25 = A | 40 | 0 | 0 | 0 |
30-40 | 35 | 36 | 10 | 1 | 36 |
40-50 | 45 | 20 | 20 | 2 | 40 |
∑ f = 140 | ∑ fd’ = 12 |
\[\overline{X}=A+\frac{\sum{f{{d}^{‘}}}}{\sum{f}}\times c\]
\[=25+\frac{12}{140}\times 10=25.85\]
Weighted Arithmetic Average
In case of simple arithmetic mean equal importance is given to every item of the series. But there may be cases where not all the items are given equal importance. So, different weights are given to the different items in accordance with the nature and purpose of the study. Weighted average is always advisable for comparative studies.
Direct Method
\[\overline{{{X}_{W}}}=\frac{\sum{Wx}}{\sum{W}}\]
Short Cut Method
\[\overline{{{X}_{W}}}=A+\frac{\sum{Wd}}{\sum{W}}\]
Step Deviation Method
\[\overline{{{X}_{W}}}=A+\frac{\sum{W{{d}^{‘}}}}{\sum{W}}\times c\]
Where,
∑ Wx = Sum of the products of the values and their corresponding weights.
∑ W = Sum of weights.
A = Assumed average.
∑ Wd = Sum of the products of deviations from the assumed average and their corresponding weights.
c = Common factor.
Example 01 |
Calculate the weighted mean of the following data:
Items | 81 | 76 | 74 | 58 | 70 | 73 |
Weights | 2 | 3 | 6 | 7 | 3 | 7 |
Solution:
From the above data we have,
x | W | Wx |
81 | 2 | 162 |
76 | 3 | 228 |
74 | 6 | 444 |
58 | 7 | 406 |
70 | 3 | 210 |
73 | 7 | 511 |
∑ W = 28 | ∑ Wx = 1961 |
\[\overline{{{X}_{W}}}=\frac{\sum{Wx}}{\sum{W}}=\frac{196}{28}=70.04\]
;