Statistics - Standard Deviation of Continuous Data Series


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

Items0-55-1010-2020-3030-40
Frequency251312

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:

Items0-1010-2020-3030-40
Frequency2113

Solution:

Based on the given data, we have:

Mean

${ \bar x = \frac{5 \times 2 + 15 \times 1 + 25 \times 1 + 35 \times 3}{7} \\[7pt] = \frac {10 + 15 + 25 + 105}{7} = 22.15 }$
ItemsMid-pt
x
Frequency
f
${\bar x}$${x-\bar x}$$f({x-\bar x})^2$
0-105222.15-17.15580.25
10-2015122.15-7.1551.12
20-3025122.152.858.12
30-4035322.1512.85495.36
  ${N=7}$  ${\sum{f(x-\bar x)^2} = 1134.85}$

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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