# Properties of Standard Deviation

## Properties of Standard Deviation

It is independent of origin

It is dependent of change of scale

It is not less than mean deviation from mean

Let n₁ and n₂ be the sizes of two series. Their means and standard deviations are x̄₁, x̄₂ and σ₁, σ₂ respectively. Let x̄ denote the combined mean of two series, that is $$\overline{x}=\frac{{{n}_{1}}{{\overline{x}}_{1}}+{{n}_{2}}{{\overline{x}}_{2}}}{{{n}_{1}}+{{n}_{2}}}$$.

∴ The combined standard deviation of two series is given by $$\sigma =\sqrt{\frac{{{n}_{1}}\left( \sigma _{1}^{2}+d_{1}^{2} \right)+{{n}_{2}}\left( \sigma _{2}^{2}+d_{2}^{2} \right)}{{{n}_{1}}+{{n}_{2}}}}$$.

Or $$\sigma =\sqrt{\frac{{{n}_{1}}\sigma _{1}^{2}+{{n}_{2}}\sigma _{2}^{2}}{{{n}_{1}}+{{n}_{2}}}+\frac{{{n}_{1}}{{n}_{2}}{{\left( {{\overline{x}}_{1}}-{{\overline{x}}_{2}} \right)}^{2}}}{{{\left( {{n}_{1}}+{{n}_{2}} \right)}^{2}}}}$$.

Where, d₁ = x̄₁ – x̄ and d₂ = x̄₂ – x̄.

$$\sigma =\frac{3\left( Quartile\,\,deviation \right)}{2}$$.

$$\sigma =\frac{5\left( Mean\,\,deviation \right)}{4}$$.

Example: The mean and standard deviation of 20 observations are found to be 10 to 2. On rechecking, it was found that an observation 8 was incorrect. Find the correct standard deviation if wrong item is omitted.

Solution: Given, x̄ = 10 and σ = 2, n = 20

∑ xᵢ/ 20 = 10

∑ xᵢ = 20 x 10

∑ xᵢ = 200

If observation 8 is omitted, then ∑ xᵢ = 200 – 8 = 192

Now, remaining number of observations = 19

Correct mean$$\,=\frac{\sum{{{x}_{i}}}}{n}=\frac{192}{19}=10.10$$.

Again, σ = 2 ⇒ σ² = 4

$$\frac{\sum{x_{i}^{2}}}{n}-{{\left( \overline{x} \right)}^{2}}=4\Rightarrow \frac{\sum{x_{i}^{2}}}{20}-{{\left( 10 \right)}^{2}}=4$$.

∑ xᵢ² = (4 + 100) x 20 = 104 x 20 = 2080

If observation 8 is omitted, then ∑ xᵢ² = 2080 – 64 = 2016.

Now, correct standard deviation $$\sigma =\sqrt{\frac{2016}{19}-{{\left( \frac{192}{19} \right)}^{2}}}$$,

$$=\sqrt{\frac{2016\times 19-{{\left( 192 \right)}^{2}}}{19\times 19}}$$,

$$=\frac{1}{19}\sqrt{38304-36864}$$,

$$=\frac{1}{19}\sqrt{1440}=\frac{37.95}{19}=1.99$$.