Statistics - Standard Deviation of Continuous Data Series

When data is given based on ranges alongwith their frequencies. Following is an example of continous series:

Items	0-5	5-10	10-20	20-30	30-40
Frequency	2	5	1	3	12

In case of continous series, a mid point is computed as $\frac{lower-limit + upper-limit}{2}$ and Standard deviation is computed using following formula.

Formula

$\sigma = \sqrt{\frac{\sum_{i=1}^n{f_i(x_i-\bar x)^2}}{N}}$

Where −

${N}$ = Number of observations = ${\sum f}$.
${f_i}$ = Different values of frequency f.
${x_i}$ = Different values of mid points for ranges.
${\bar x}$ = Mean of mid points for ranges.

Example

Problem Statement:

Let's calculate Standard Deviation for the following continous data:

Items	0-10	10-20	20-30	30-40
Frequency	2	1	1	3

Solution:

Based on the given data, we have:

Mean

${ \bar x = \frac{5 \times 2 + 15 \times 1 + 25 \times 1 + 35 \times 3}{7} \\[7pt] = \frac {10 + 15 + 25 + 105}{7} = 22.15 }$

Items	Mid-pt x	Frequency f	${\bar x}$	${x-\bar x}$	$f({x-\bar x})^2$
0-10	5	2	22.15	-17.15	580.25
10-20	15	1	22.15	-7.15	51.12
20-30	25	1	22.15	2.85	8.12
30-40	35	3	22.15	12.85	495.36
		${N=7}$			${\sum{f(x-\bar x)^2} = 1134.85}$

Based on the above mentioned formula, Standard Deviation $ \sigma $ will be:

${ \sigma =\sqrt{\frac{\sum_{i=1}^n{f_i(x_i-\bar x)^2}}{N}} \\[7pt] \, = \sqrt{\frac{1134.85}{7}} \, = 12.73}$

The Standard Deviation of the given numbers is 12.73.