# Geometric Mean

The Geometric Mean (G.M) of a series containing n observations is the nth root of the product of the values.

Consider, if x1 , x2 …. Xn are the observation, then the G.M is defined as

$G. M = \sqrt[n]{x_{1},x_{2},…x_{n}}$ $G. M = (x_{1},x_{2},…x_{n})^{^{\frac{1}{n}}}$

This can also be written as

Log GM = $\frac{1}{n}\log (x_{1},x_{2}….x_{n})$

=$\frac{1}{n}(\log x_{1}+\log x_{2}+….+\log x_{n})$

=$\frac{\sum \log x_{i}}{n}$

Therefore, Geometric Mean, GM = $Antilog\frac{\sum \log x_{i}}{n}$

Where n = f1 + f2 +…..+ fn

It is also represented as,

$\sqrt[n]{\prod_{i=1}^{n}x_{i}}$

For any Grouped Data, G.M can be written as

GM = $Antilog\frac{\sum f \log x_{i}}{n}$

There is a difference in arithmetic mean(mean) and geometric mean. Now let us see the difference between them.

## Geometric Mean Vs Arithmetic Mean

There is a difference between both the means like G.M and arithmetic mean for the given data set how they are calculations are done using G.M and arithmetic mean formula.

 Arithmetic Mean Geometric Mean The arithmetic mean or mean can be found by adding all the numbers for the given data set divided by the number of data points in a set. It can be found by multiplying all the numbers in the given data set and take the nth root for the obtained result. For example, the given data sets are: 5, 10, 15 and 20 Here, the number of data points = 4 Arithmetic mean or mean = (5+10+15+20)/4 Mean = 50/4 =12.5 For example, consider the given data set, 4, 10, 16, 24 Here n= 4 Therefore, the G.M = 4th root of (4 ×10 ×16 × 24) = 4th root of 15360 G.M = 11.13

## Geometric Mean Properties

Some of the important properties of the G.M are:

• The G.M for the given data set is always less than the arithmetic mean for the data set
• If each object in the data set is substituted by the G.M, then the product of the objects remains unchanged.
• The ratio of the corresponding observations of the G.M in two series is equal to the ratio of their geometric means
• The products of the corresponding items of the G.M in two series are equal to the product of their geometric mean.

## Application of Geometric Mean

The greatest assumption of the G.M is that data can be really interpreted as scaling factor. Before that, we have to know when to use the G.M . The answer to this is , it should be only applied to positive values and often used for the set of numbers whose values are exponential in nature and whose values are meant to be multiplied together. This means that there will be no zero value and negative value which we cannot really apply. Geometric mean has lot of advantages and it is used in many fields. Some of the applications are as follows

• It is used in stock indexes. Because many of the value line indexes which is used by financial departments use G.M.
• It is used to calculate the annual return on portfolio.
• It is used in finance to find the average growth rates which also referred as compounded annual growth rate.
• It is also used in studies like cell division and bacterial growth etc.

## Geometric Mean Problems

Here you are provided with geometric mean examples as follows

### Question 1 :

Find the G.M of the values 10, 25, 5, and 30

### Solution :

Given 10, 25, 5, 30

We know that,

GM = $\sqrt[n]{\prod_{i=1}^{n}x_{i}}$

= $\sqrt[4]{10\times 25\times 5\times 30}$

= $\sqrt[4]{37500}$

= 13.915

Therefore, the geometric mean = 13.915

### Question 2 :

Find the geometric mean of the following data.

 Weight of ear head x ( g) Log x 45 1.653 60 1.778 48 1.681 100 2.000 65 1.813 Total 8.925

### Solution:

Here n=5

GM =  $Antilog\frac{\sum \log x_{i}}{n}$

= Antilog 8.925/5

= Antilog 1.785

= 60.95

Therefore the G.M of the given data is 60.95

### Question 3 :

Find the geometric mean of the following grouped data for the frequency distribution of weights.

 Weights of ear heads (g) No of ear heads (f) 60-80 22 80-100 38 100-120 45 120-140 35 140-160 20 Total 160

### Solution :

 Weights of ear heads (g) No of ear heads (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 73.99 140-160 20 150 2.716 43.52 Total 160 324.2

From the given data, n = 160

We know that the G.M for the grouped data is

GM = $Antilog\frac{\sum f \log x_{i}}{n}$

GM = Antilog ( 324.2 /160 )

GM = Antilog ( 2.02625 )

GM = 106.23

Therefore, the G.M = 106.23

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