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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