# Combined Variance

Like combined mean, the combined variance or standard deviation can be calculated for different sets of data. Suppose we have two sets of data containing $${n_1}$$ and $${n_2}$$ observations with means $${\overline X _1}$$ and $${\overline X _2}$$ and variances $${S_1}^2$$ and $${S_2}^2$$. If $${\overline X _c}$$ is the combined mean and $${S_c}^2$$ is the combined variance of $${n_1} + {n_2}$$ observations, then combined variance is given by:

\[{S_c}^2 = \frac{{{n_1}{S_1}^2 + {n_2}{S_2}^2 + {n_1}{{\left( {{{\overline X }_1} – {{\overline X }_c}} \right)}^2} + {n_2}{{\left( {{{\overline X }_2} – {{\overline X }_c}} \right)}^2}}}{{{n_1} + {n_2}}}\]

It can be written as:

\[{S_c}^2 = \frac{{{n_1}\left[ {{S_1}^2 + {{\left( {{{\overline X }_1} – {{\overline X }_c}} \right)}^2}} \right] + {n_2}\left[ {{S_2}^2 + {{\left( {{{\overline X }_2} – {{\overline X }_c}} \right)}^2}} \right]}}{{{n_1} + {n_2}}}\]

Here \[{X_c} = \frac{{{n_1}{{\overline X }_1} + {n_2}{{\overline X }_2}}}{{{n_1} + {n_2}}}\]

The combined standard deviation $${S_c}$$ can be calculated by taking the square root of $${S_c}^2$$.

__Example__:

For a group of 50 male workers the mean and standard deviation of their daily wages are 63 dollars and 9 dollars respectively. For a group of 40 female workers these values are 54 dollars and 6 dollars respectively. Find the mean and variance of the combined group of 90 workers.

__Solution__:

Here $${n_1} = 50$$, $${\overline X _1} = 63$$, $${S_1}^2 = 81$$

$${n_2} = 40$$, $${\overline X _2} = 54$$, $${S_2}^2 = 36$$

Combined Arithmetic Mean $$ = {X_c} = \frac{{{n_1}{{\overline X }_1} + {n_2}{{\overline X }_2}}}{{{n_1} + {n_2}}}$$

$$ = {X_c} = \frac{{50\left( {63} \right) + 40\left( {54} \right)}}{{50 + 40}} = \frac{{5310}}{{90}} = 59$$

Combined Variance $${S_c}^2 = \frac{{{n_1}\left[ {{S_1}^2 + {{\left( {{{\overline X }_1} – {{\overline X }_c}} \right)}^2}} \right] + {n_2}\left[ {{S_2}^2 + {{\left( {{{\overline X }_2} – {{\overline X }_c}} \right)}^2}} \right]}}{{{n_1} + {n_2}}}$$

$$ = \frac{{50\left[ {81 + {{\left( {63 – 59} \right)}^2}} \right] + 40\left[ {36 + {{\left( {54 – 59} \right)}^2}} \right]}}{{50 + 40}}$$

$$ = \frac{{4850 + 2440}}{{90}} = \frac{{7290}}{{90}} = 81$$

Ati

December 16@ 10:55 pmHello,

Thank you for this clear and nice post.

May I have the general formula for more than 2 groups with unequal sample size?

May I know the difference between the Combined variance and Pooled variance?

Sincerely,

Ati