Two sample of size 100 and 150 respectively have means 50 and 60 deviation of the combined sample of size 250.

Explanation

`{:("Given":,N_(1)=100",",barX_(1)=50",",sigma_(1)=5),(,N_(2)=150",",barX_(2)=60",",sigma_(2)=6):}`
Now,
`barX_(12)=(N_(1)barX_(1)+N_(2)barX_(2))/(N_(1)+N_(2))`
`=(100xx50+150xx60)/(100+150)=(5,000+9,000)/(250)`
`(14,000)/(250)=56`
`d_(1)=barX_(1)-barX_(12)=50-56=-6`
`d_(2)=barX_(2)-barX_(12)=60-56=+4`
`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)))`
`=sqrt((100xx(5)^(2)+150xx(6)^(2)+100xx(-6)^(2)+150xx(4)^(2))/(100+150))`
`=sqrt((100xx25+150xx36+100xx36+150xx16)/(250))`
`=sqrt((2,500+5,400+3,600+2,400)/(250))=sqrt((13,900)/(250))`
`=sqrt(55.6)=7.46`
Hence, the Combined Mean =56 and the Combined Standard Deviation=7.46.