Geometric mean for grouped data
Let $(x_i,f_i), i=1,2, \cdots , n$ be the given frequency distribution then the geometric mean of $X$ is denoted by $GM$.
Formula
The geometric mean of grouped data is given by
$GM=\bigg(\prod_{i=1}^n x_i^{f_i}\bigg)^{1/N}$
OR
$\log (GM) =\frac{1}{N}\sum_{i=1}^n f_i\log(x_i)$
where,
$N=\sum_i f_i$
total number of observations
Example
Compute geometric mean for the following frequency distribution.
x | 5-8 | 9-12 | 13-16 | 17-20 | 21-24 |
---|---|---|---|---|---|
f | 2 | 13 | 21 | 14 | 5 |
Solution
Class Interval | Class Boundries | mid-value ($x_i$) | Freq ($f_i$) | $log(x_i)$ | $f_i*log(x_i)$ | |
---|---|---|---|---|---|---|
5-8 | 4.5-8.5 | 6.5 | 2 | 1.8718 | 3.7436 | |
9-12 | 8.5-12.5 | 10.5 | 13 | 2.3514 | 30.5682 | |
13-16 | 12.5-16.5 | 14.5 | 21 | 2.6741 | 56.1561 | |
17-20 | 16.5-20.5 | 18.5 | 14 | 2.9178 | 40.8492 | |
21-24 | 20.5-24.5 | 22.5 | 5 | 3.1135 | 15.5675 | |
Total | 55 | 146.8846 |
The log of geometric mean is
$$ \begin{aligned} \log (GM) &=\frac{1}{N}\sum_{i=1}^n f_i\log(x_i)\\ &=\frac{146.8846}{55}\\ &=2.6706 \end{aligned} $$
The geometric mean is
$$ \begin{aligned} GM & =\exp(\log (GM))\\ &=\exp(2.6706)\\ &=14.4486 \end{aligned} $$