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.
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