For a group containing 100 observations, the arithmetic mean and standard deviation are 8 and `sqrt(10.5)`. For 50 observations selected from the 100

Monjur Alahi

For a group containing 100 observations, the arithmetic mean and standard deviation are 8 and `sqrt(10.5)`. For 50 observations selected from the 100 observations, the arithmetic mean and standard deviations are 10 and 2 respectively. Find the arithmetic mean and the standard deviation of the other half.


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Salim Ahmed Kawsar

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Given : N=100,`barX_(12)`=8, `sigma_(12)=sqrt(10.5)`
`N_(1)`=50, `barX_(1)`=10, `sigma_(1)`=2
`N_(2)=100-N_(1)`=100-50=50
we know that,
`barX_(12)=(N_(1)barX_(1)+N_(2)barX_(2))/(N_(1)+N_(2))`
`8=(50(10)+50(barX_(2)))/(100)`
`800=500+50barX_(2)`
`50barX_(2)=800-500`
`50barX_(2)=300`
`barX_(2)=(300)/(500)=6`
`d_(1)=barX_(1)-barX_(12)=10-8=2`
`d_(2)=barX_(2)-barX_(12)=6-8=-2`
`sigma_(12)=sqrt(N_(1)sigma_(1)^(2)+N_(2)sigma_(2)^(2)+N_(1)d_(1)^(2)+N_(2)d_(2)^(2))/(N_(1)+N_(2))`
Substituting the values, we get
`sqrt(10.5) =sqrt(50xx(2)^(2)+50sigma_(2)^(2)+50xx(2)^(2)+50xx(-2)^(2))/(100)`
`sqrt(10.5) =sqrt(50xx4+50sigma_(2)^(2)+50xx4+50xx4)/(100)`
Squaring both sides,
`10.5=(200+50sigma_(2)^(2)+200+200)/(100)`
`10.5xx100=600+50sigma_(2)^(2)`
`1,050=600+50sigma_(2)^(2)`
`:. " " 50sigma_(2)^(2)=1,050-600`
`:. " " 50sigma_(2)^(2)=450`
`:. " " sigma_(2)^(2)=(450)/(50)=9`
`implies " " sigma_(2)=3`
Thus, `barX_(2)`=6, `sigma_(2)`=3.

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