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Worked Examples · Example 9

Q.Two groups of items have the following statistics. Group 1: 4040 items, mean 3030, standard deviation 44. Group 2: 6060 items, mean 3535, standard deviation 55. Find the combined mean and the combined standard deviation of all 100100 items.

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Here n1=40n_1 = 40, xˉ1=30\bar{x}_1 = 30, σ1=4\sigma_1 = 4; and n2=60n_2 = 60, xˉ2=35\bar{x}_2 = 35, σ2=5\sigma_2 = 5.

Combined mean:

xˉ12=n1xˉ1+n2xˉ2n1+n2=40(30)+60(35)40+60=1200+2100100=3300100=33.\bar{x}_{12} = \frac{n_1\bar{x}_1 + n_2\bar{x}_2}{n_1 + n_2} = \frac{40(30) + 60(35)}{40 + 60} = \frac{1200 + 2100}{100} = \frac{3300}{100} = 33.

Deviations of the group means from the combined mean:

d1=xˉ1−xˉ12=30−33=−3,d2=xˉ2−xˉ12=35−33=2.d_1 = \bar{x}_1 - \bar{x}_{12} = 30 - 33 = -3, \qquad d_2 = \bar{x}_2 - \bar{x}_{12} = 35 - 33 = 2.

Combined standard deviation:

σ12=n1(σ12+d12)+n2(σ22+d22)n1+n2=40(42+(−3)2)+60(52+22)100.\sigma_{12} = \sqrt{\frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}} = \sqrt{\frac{40(4^2 + (-3)^2) + 60(5^2 + 2^2)}{100}}.

Evaluate inside: 40(16+9)+60(25+4)=40(25)+60(29)=1000+1740=274040(16 + 9) + 60(25 + 4) = 40(25) + 60(29) = 1000 + 1740 = 2740. So

σ12=2740100=27.4≈5.23.\sigma_{12} = \sqrt{\frac{2740}{100}} = \sqrt{27.4} \approx 5.23. …

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