The mean and standard deviation of a group of 100 observations were found to be 20 and 3 respectively. Later on it was found that three observations were incorrect which were recorded as 21,21 and 18. Find the mean and standard deviation if the incorrect observation are omitted
Given, number of observation n=100
Incorrect mean (¯x)=20
We know that
¯x=∑xin
⇒20=1100100∑i=1xi
⇒100∑i=lxi=20×100=2000
So, the incorrect sum of observations =2000.
Given incorrect observations are 21,21,18 and these has to be omitted.
So, correct sum of observations =2000−21−21−18=2000−60=1940
Correct mean =Correctsum100−3=194097=20
Given, incorrect standard deviation σ=3
σ=
⎷∑ni=1x2in−(∑ni=1xin)2
σ2=∑ni=1x2in−(¯x)2
⇒32=∑x2i100−(20)2
⇒Incorrect∑x2i=100(9+400)=40900
Correct n∑i=lx2i=Incorrectn∑i=lx2i−(21)2−(21)2−(18)2
=40900−441−441−324
=39694
∴ Correct standard deviation = √correct∑x2in−(Correctmean)2
=√3969497−(20)2
=√409.216−400
=√9.216
=3.036