Q. The mean deviation of items $x$, $x + y$, $x + 2y$, $...$, $x + 2ny$ from mean is
Statistics
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 $= 2n + 1$)
$x_i$
$d = |x_i - \bar{x}|$
$x$
$ny$
$x+y$
$(n-1)y$
$x+2y$
$(n-2)y$
$x+3y$
$(n-3)y$
$⋮$
$⋮$
$⋮$
$⋮$
$x + ny$
$0$
$x + (n + 1 )y$
$y$
$⋮$
$⋮$
$⋮$
$⋮$
$x + 2ny$
$ny$
$\Sigma \left|d\right| = 2 \left[ny + \left(n - \right)y + ... +y\right]$
$= \frac{2\cdot y\,n\left(n+1\right)}{2} = ny\left(n+1\right)$
$\therefore M.D. \left(\bar{x}\right) = \frac{\Sigma \left|d\right|}{2n+1}$
$= \frac{ny\left(n+1\right)}{2n+1}$
| $x_i$ | $d = |x_i - \bar{x}|$ |
|---|---|
| $x$ | $ny$ |
| $x+y$ | $(n-1)y$ |
| $x+2y$ | $(n-2)y$ |
| $x+3y$ | $(n-3)y$ |
| $⋮$ | $⋮$ |
| $⋮$ | $⋮$ |
| $x + ny$ | $0$ |
| $x + (n + 1 )y$ | $y$ |
| $⋮$ | $⋮$ |
| $⋮$ | $⋮$ |
| $x + 2ny$ | $ny$ |