Geometric Mean

The geometric mean is another measure of central tendency based on mathematical footing, like arithmetic mean. The geometric mean can be defined as:

“The geometric mean is the nth positive root of the product of ‘n’ positive given values.”

Hence, the geometric mean for a value $$X$$ containing $$n$$ values such as $${x_1},{x_2},{x_3},…,{x_n}$$ is denoted by $$G.M$$ of $$X$$ and given as:

\[G.M{\text{ of }}X = \overline X = \sqrt[n]{{{x_1} \cdot {x_2} \cdot {x_3} \cdot \cdots \cdot {x_n}}}\] (for ungrouped data)

If we have a series of $$n$$ positive values with repeated values such as $${x_1},{x_2},{x_3},…,{x_k}$$ which are repeated $${f_1},{f_2},{f_3},…,{f_k}$$ times respectively, then the geometric mean will become:

\[G.M{\text{ of }}X = \overline X = \sqrt[n]{{{x_1}^{{f_1}} \cdot {x_2}^{{f_2}} \cdot {x_3}^{{f_3}} \cdot \cdots \cdot {x_k}^{{f_k}}}}\] (For Grouped Data)

Where $$n = {f_1} + {f_2} + {f_3} + \cdots + {f_k}$$

 

Example:

Find the geometric mean of the values 10, 5, 15, 8, 12.

Solution:
Here$${x_1} = 10$$, $${x_2} = 5$$$${x_3} = 15$$$${x_4} = 8$$$${x_5} = 12$$ and $$n = 5$$

$$G.M{\text{ of }}X = \overline X = \sqrt[5]{{10 \times 5 \times 15 \times 8 \times 12}}$$
$$\overline X = \sqrt[5]{{72000}} = {(72000)^{\frac{1}{5}}} = 9.36$$

 

Example:

Find the geometric mean of the following data:

$$X$$
$$13$$
$$14$$
$$15$$
$$16$$
$$17$$
$$f$$
$$2$$
$$5$$
$$13$$
$$7$$
$$3$$

Solution:
We may write it as below:

Here $${x_1} = 13$$, $${x_2} = 14$$$${x_3} = 15$$$${x_4} = 16$$$${x_5} = 17$$

$${f_1} = 2$$, $${f_2} = 5$$, $${f_3} = 13$$, $${f_4} = 7$$, $${f_5} = 3$$

$$n = \sum f = {f_1} + {f_2} + {f_3} + {f_4} + {f_5} = 2 + 5 + 13 + 7 + 3 = 30$$

Using the formula of geometric mean for grouped data, the geometric mean in this case will become:

\[G.M{\text{ of }}X = \overline X = \sqrt[n]{{{x_1}^{{f_1}} \cdot {x_2}^{{f_2}} \cdot {x_3}^{{f_3}} \cdot {x_4}^{{f_4}} \cdot {x_5}^{{f_5}}}}\]

\[\overline X = \sqrt[{30}]{{{{(13)}^2} \cdot {{(14)}^5} \cdot {{(15)}^{13}} \cdot {{(16)}^7} \cdot {{(17)}^3}}}\]

\[\overline X = \sqrt[{30}]{{2.33292 \times {{10}^{35}}}} = {(2.33292 \times {10^{35}})^{\frac{1}{{30}}}}\]

\[\overline X = 15.0984 \approx 15.10\]

The method explained above to calculate the geometric mean is useful when the values in the given data are small in number and an electronic calculator is available. When a set of data contains a large number of values then we need an alternate way to compute the geometric mean. The modified or alternative way of computing the geometric mean is given as:

For Ungrouped Data
For Grouped Data
$$G.M{\text{ of }}X = \overline X = Anti\log \left( {\frac{{\sum \log x}}{n}} \right)$$
$$G.M{\text{ of }}X = \overline X = Anti\log \left( {\frac{{\sum f\log x}}{{\sum f}}} \right)$$

 

Example:
Find the geometric mean of the values 10, 5, 15, 8, 12

$$x$$
$$\log x$$
$$10$$
$$1.0000$$
$$5$$
$$0.6990$$
$$15$$
$$1.1761$$
$$8$$
$$0.9031$$
$$12$$
$$1.0792$$
Total
$$\sum \log x = 4.8573$$

$$G.M{\text{ of }}X = \overline X = Anti\log \left( {\frac{{\sum \log x}}{n}} \right)$$
$$\overline X = Anti\log \left( {\frac{{4.8573}}{5}} \right)$$
$$\overline X = Anti\log \left( {0.9715} \right)$$
$$\overline X = 9.36$$

 

Example:

Find the geometric mean for the following distribution of students’ marks:

Marks
$$0 – 30$$
$$30 – 50$$
$$50 – 80$$
$$80 – 100$$
No. of Students
$$20$$
$$30$$
$$40$$
$$10$$

 

Solution:

Marks
No. of Students
$$f$$
Mid Points
$$x$$
$$f\log x$$
$$0 – 30$$
$$20$$
$$15$$
$$20\log 15 = 23.5218$$
$$30 – 50$$
$$30$$
$$40$$
$$30\log 40 = 48.0168$$
$$50 – 80$$
$$40$$
$$65$$
$$40\log 65 = 72.5165$$
$$80 – 100$$
$$10$$
$$90$$
$$10\log 90 = 19.5424$$
Total
$$\sum f = 100$$
 
$$\sum f\log x = 163.6425$$

$$G.M{\text{ of }}X = \overline X = Anti\log \left( {\frac{{\sum f\log x}}{{\sum f}}} \right)$$
$$\overline X = Anti\log \left( {\frac{{163.6425}}{{100}}} \right)$$
$$\overline X = Anti\log \left( {1.6364} \right)$$
$$\overline X = 43.29$$