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 $n = 5$

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

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 $\sum f = 100$ $\sum f\log x = 163.6425$

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