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Q.

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

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a

64

b

65.2

c

67.2

d

64.2

answer is C.

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Detailed Solution

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$\bar{x} = \frac{n_{1} {\bar{x}}_{1} + n_{2} {\bar{x}}_{2}}{n_{1} + n_{2}} = \frac{200 \times 25 + 300 \times 10}{500} = 16 d_{1} = {\bar{x}}_{1} - \bar{x} = 9, d_{2} = {\bar{x}}_{2} - \bar{x} = - 6 σ^{2} = \frac{n_{1} (σ_{1}^{2} + d_{1}^{2}) + n_{2} (σ_{2}^{2} + d_{2}^{2})}{n_{1} + n_{2}} = 67.2$

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