Statistics - Standard Deviation of Discrete Data Series
When data is given alongwith their frequencies. Following is an example of discrete series:
| Items | 5 | 10 | 20 | 30 | 40 | 50 | 60 | 70 |
|---|---|---|---|---|---|---|---|---|
| Frequency | 2 | 5 | 1 | 3 | 12 | 0 | 5 | 7 |
For discrete series, the Standard Deviation can be calculated using the 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 variable x.
Example
Problem Statement:
Calculate Standard Deviation for the following discrete data:
| Items | 5 | 15 | 25 | 35 |
|---|---|---|---|---|
| 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 x | Frequency f | ${\bar x}$ | ${x-\bar x}$ | $f({x-\bar x})^2$ |
|---|---|---|---|---|
| 5 | 2 | 22.15 | -17.15 | 580.25 |
| 15 | 1 | 22.15 | -7.15 | 51.12 |
| 25 | 1 | 22.15 | 2.85 | 8.12 |
| 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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