The mean of two samples of sizes 200 and 300 was found to be 25, and 10 respectively. Their standard deviations were 3 and 4 respectively. The variance of a combined sample of size 500 is

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64.2

65.2

67.2

answer is C.

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Detailed Solution

Combined mean $\bar{x} = \frac{n_{1} \bar{x_{1}} + n_{2} \bar{x_{2}}}{n_{1 +} n_{2}} = \frac{200 \times 25 + 300 \times 10}{500} = 16$

Let $d_{1} = \bar{x_{1}} - \bar{x} = 25 - 16 = 9 a n d d_{2} = \bar{x_{2}} - \bar{x} = 10 - 16 = - 6$

Now $σ^{2} = \frac{n_{1} (σ_{1}^{2} + d_{1}^{2}) + n_{2} (σ_{2}^{2} + d_{2}^{2})}{n_{1} + n_{2}} = \frac{200 (9 + 81) + 300 (16 + 36)}{500} = \frac{33600}{500} = 67.2$

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