If `barx` is the mean of n observations `x_(1),x_(2),x_(3)……x_(n)`, then the value of `sum_(i=1)^(n)(x_(i)-barx)` is (i) `-1` (ii) 0 (iii) 1 (iv) n-1

Options

A. `S le r sqrt((n)/(n-1))`
B. `S=r sqrt((n)/(n-1))`
C. `S ge r sqrt((n)/(n-1))`
D. None of these

Answer

A

Explanation

We have r=max `|x_(i)-x_(j)|`
and `S^(2)=(1)/(n-1) underset(i=1)overset(n)(sum)(x_(i)-overline(x))^(2)`
Now, `(x_(i)-overline(x))^(2)=(x_(i)-(x_(1)+x_(2)+..+x_(n))/(n))^(2)`
`=(1)/(n^(2))[(x_(i)-x_(1))+(x_(i)-x_(2))+..+(x_(i)-x_(i)-1)+(x_(i)-x_(1)+1)+..+(x_(i)-x_(n))] le (1)/(n^(2))[(n-1)r]^(2) " " [because]x_(i)-x_(j)|le r|`
`(x_(i)-overline(x))^(2) le r^(2)`
`implies underset(i=1)overset(n)(sum)(x_(i)overline(x))^(2) le nr^(2)`
`implies (1)/(n-1) underset(i=1)overset(n)(sum)(x_(i)-overline(x))^(2) le(nr^(2))/((n-1))`
`implies S^(2) le (nr^(2))/((n-1))`
`implies S le r sqrt((n)/(n-1))`