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InterviewSolution
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The mean and standard deviation of 100 observations were calculated as40 and 5.1 respectively by a student who took by mistake 50 instead of 40 forone observation. What are the correct mean sand standard deviation? |
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Answer» `"Mean "=40rArr(overset(100)underset(i=1)Sigmax_(i))/(10)=40` `rArr" "overset(100)underset(i=1)Sigmax_(i)=4000.` `SD=5.1rArrsigma^(2)=(5.1)^(2)` `rArr" "(overset(100)underset(i=1)Sigmax_(i)^(2))/(100)-(40)^(2)=26.01` `rArr" "overset(100)underset(i=1)Sigmax_(i)^(2)=162601.` Thus, incorrect `(overset(100)underset(i=1)Sigmax_(i))=4000 and"incorrect "(overset(100)underset(i=1)Sigmax_(i)^(2))=162601` Now, incorrect `(overset(100)underset(i=1)Sigmax_(i))=4000` `rArr" correct "(overset(100)underset(i=1)Sigmax_(i))=(4000-50+40)=3990` `rArr" correct mean"=(3990)/(100)=39.9." ...(i)"` `"And, correct "(overset(100)underset(i=1)Sigmax_(i)^(2))=162601` `rArr" correct"(overset(100)underset(i=1)Sigmax_(i)^(2))={162601-(50)^(2)+(40)^(2)}=161701` `rArr" correct variance "=("correct "(overset(100)underset(i=1)Sigmax_(i)^(2)))/(100)-("correct mean")^(2)` `={(161701)/(100)-(39.9)^(2)}={1617.01-(40-0.1)^(2)}` `=(1617.01)-{1600+0.01-8}` `=(1617.01-1592.01)=25` `rArr" correct "SD=sqrt(25)=5.` Hence, correct mean =39.9 and correct SD=5. |
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