1. Tardigrade
  2. Question
  3. Mathematics
  4. The mean deviation of items x, x + y, x + 2y, ..., x + 2ny from mean is

Q. The mean deviation of items $x$, $x + y$, $x + 2y$, $...$, $x + 2ny$ from mean is

Statistics Report Error

Solution:

We need mean deviation about the mean,
$\therefore \bar{x} = \frac{x+\left(x+y\right)+\left(x+2y\right)+......+\left(x+2ny\right)}{2n+1}$
$= \frac{\left(2n+1\right)}{2} \frac{\left[x+x+2ny\right]}{\left(2n+1\right)} = x+ny$
(Total number of observations $= 2n + 1$)
$x_i$ $d = |x_i - \bar{x}|$
$x$ $ny$
$x+y$ $(n-1)y$
$x+2y$ $(n-2)y$
$x+3y$ $(n-3)y$
$⋮$ $⋮$
$⋮$ $⋮$
$x + ny$ $0$
$x + (n + 1 )y$ $y$
$⋮$ $⋮$
$⋮$ $⋮$
$x + 2ny$ $ny$

$\Sigma \left|d\right| = 2 \left[ny + \left(n - \right)y + ... +y\right]$
$= \frac{2\cdot y\,n\left(n+1\right)}{2} = ny\left(n+1\right)$
$\therefore M.D. \left(\bar{x}\right) = \frac{\Sigma \left|d\right|}{2n+1}$
$= \frac{ny\left(n+1\right)}{2n+1}$

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