Question
A set of four observations has mean 1 and variance 13. Another set of six observations has mean 2 and variance 1. Then, the variance of all these 10 observations is equal to :
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Solution
Sol. [Key Concept: Combined variance $= \frac{n_{1}\sigma_{1}^{2} + n_{2}\sigma_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1}n_{2}}{(n_{1} + n_{2})^{2}}(m_{1} - m_{2})^{2} = 6.04$, where $m_{1},m_{2}$ are the two group means.]
Given $\overline{x} = 1,\sigma_{1}^{2} = 13$
$$\overline{y} = 2,\sigma_{2}^{2} = 1 $$Combined variance $= \frac{n_{1}\sigma_{1}^{2} + n_{2}\sigma_{2}^{2}}{n_{1} + n_{2}} + \frac{n_{1}n_{2}}{\left( n_{1} + n_{2} \right)^{2}}(\overline{x} - \overline{y})^{2}$
$${= \frac{4 \times 13 + 6 \times 1}{10} + \frac{4 \times 6}{10^{2}}(2 - 1)^{2} }{\frac{58}{10} + \frac{24}{100} = \frac{580 + 24}{100} = \frac{604}{100} = 6.04 }$$
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