Arithmetic Mean formula With Example


Statistics / Friday, August 3rd, 2018

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:

Types of Arithmetic Mean

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:

Arithmetic mean formula with example_method

Direct Method

Individual Series

Let X is the variable which takes values x1, x2, x3, …, xn 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.

MonthJanFebMarAprMayJune
Income($)250003000045000200002500020000

Solution:

n = Total number of items (observations) = 6
Total income = ∑ xi = $(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

Wage1020304050
No. of Workers45325

Solution:

Let us denote the wages by x and number of workers by f.

Wages (x)

No. of Workers (f)fx

10

440

20

5100

30

3

90

402

80

505

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.

Wage110112113117125128130
No. of Workers2517131514862

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

252750
11217

1904

113

131469
11715

1755

a

1414a
1258

1000

128

6768
1302

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 Interval84-9090-9696-102102-108108-114
frequency81215105

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
84-90978

776

90-96

93121116
96-1029915

1485

102-108

105101050
108-1141115

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 = xi – 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 ($)22002500300037004500
No. of Staff5101573

Solution:
Let assumed mean A = 3000

xfD = (x – A)fd
22005-800-4000
250010-500-5000
300015000000
370077004900
4500315004500
∑ 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 Marks0-1010-2020-3030-4040-50
No. of students5372520

Solution:
Let A = 25

Marks of Students

No. of Students(f)Mid-values(m)d = m – 25

fd

0-10

55-20-100
10-20315-10

-30

20-30

725000
30-40253510

250

40-50

204520400
∑ 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

Marks0-1010-2020-3030-4040-50
No. of students2024403620

Solution:

Let Assumed average A = 25

Marks

Mid-values(m)fA = 25,

d = m – 25

d’ = (d/10)

fd’

0-10

520-20-2-40
10-201524-10-1

-24

20-30

25 = A40000
30-403536101

36

40-50

452020240
∑ 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:

Items817674587073
Weights236737

Solution:

From the above data we have,

x

WWx
812

162

76

3228
746

444

58

7406
703

210

73

7511
∑ W = 28

∑ Wx = 1961

\[\overline{{{X}_{W}}}=\frac{\sum{Wx}}{\sum{W}}=\frac{196}{28}=70.04\]

 

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