# Arithmetic Mean formula With Example

Statistics / Friday, August 3rd, 2018
(Last Updated On: August 6, 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:

## 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 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.

 Month Jan Feb Mar Apr May June Income($) 25000 30000 45000 20000 25000 20000 Solution: n = Total number of items (observations) = 6 Total income = ∑ xi =$(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  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 = 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 ($) 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

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