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Question
Mean and standard deviation of two distributions of 100 and 150 items are 50, 5 and 40, 6 respectively. Find the mean and standard deviations of all the 250 items taken together.
Solution
Here, n1 = 100, n2 = 150, `bar"x"_1` = 50, `bar"x"_2 = 40`, `sigma_1` = 5, `sigma_2` = 6
Combined Mean is given by,
`bar"x"_"c" = ("n"_1bar"x"_1 + "n"_2bar"x"_2)/("n"_1 + "n"_2)`
= `((100)(50) + (150)(40))/(100 + 150)`
= `(5000 + 6000)/(250)`
= `(11000)/(250)`
= 44
Combined standard deviation is given by,
`sigma_"c" = sqrt(("n"_1 (sigma_1^2 + "d"_1^2) + "n"_2 (sigma _2^2 + "d"_2^2))/("n"_1 + "n"_2)`
where d1 = `bar"x"_1-bar"x"_"c"`, d2 = `bar"x"_2-bar"x"_"c"`
∴ d1 = 50 − 44 = 6 and d2 = 40 − 44 = −4
∴ σc = `sqrt((100(5^2 + 6^2) + 150(6^2 + (-4)^2))/(100 + 150)`
= `sqrt((100(25 + 36) + 150(36 + 16))/(250)`
= `sqrt((100(61) + 150(52))/(250)`
= `sqrt((6100 + 7800)/(250)`
= `sqrt(13900/250)`
= `sqrt(55.6)`
∴ The mean and standard deviation of all 250 items taken together are 44 and `sqrt55.6` respectively.
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