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If both the standard deviation and mean of data `x_(1),x_(2),x_(3),……x_(50)` are 16, then the mean of the data set (`x_(1)-4)^(2),(x_(2)-4)^(2),(x_(3)-4)^(2),….(x_(50)-4)^(2)` isA. 480B. 400C. 380D. 525 |
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Answer» Correct Answer - B It is given that both mean and standard deviation of 50 observations `x_(1), x_(2), x_(3), … , x_(50)` are equal to 16, So, mean `=(Sigma x_(i))/(50)=16 " …(i)" ` and standard deviation `=sqrt((Sigmax_(i)^(2))/(50)-((Sigmax_(i))/(50))^(2))=16` `rArr (Sigmax_(i)^(2))/(50)-(16)^(2)=(16)^(2)` `rArr (Sigmax_(i)^(2))/(50)=2 xx 256 = 512 " ...(ii)" ` Now, mean of `(x_(1)-4)^(2),(x_(2)-4)^(2), ..., (x_(50)-4)^(2)` `=(Sigma(x_(i)-4)^(2))/(50)=(Sigma(x_(i)^(2)-8x_(i)+16))/(50)` `=(Sigmax_(i)^(2))/(50)-8((Sigmax_(i))/(50))+(16)/(50)Sigma1` `=512-(8xx16)+((16)/(50)xx50) " " `[from Eqs. (i) and (ii)] `=512 -128 +16 =400` |
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