The probability distribution of X is:

$X$	0	1	2	3
$P\left(X\right)$	$\frac{1-d}{4}$	$\frac{1+2d}{4}$	$\frac{1-4d}{4}$	$\frac{1+3d}{4}$

For the minimum possible value of $d,$ 120 times the mean of $X$ is equal to___________.

$0\le \frac{1-d}{4}\le 1\Rightarrow -3\le d\le 1\mathrm{..........}\left(i\right)$

$0\le \frac{1+2d}{4}\le 1\Rightarrow -\frac{1}{2}\le d\le \frac{3}{2}\mathrm{......}\left(ii\right)$

$0\le \frac{1-4d}{4}\le 1\Rightarrow -\frac{3}{4}\le d\le \frac{1}{4}\mathrm{.....}\left(iii\right)$

$0\le \frac{1+3d}{4}\le 1\Rightarrow -\frac{1}{3}\le d\le 1\mathrm{.......}\left(iv\right)$

From $\left(i\right),\left(ii\right),\left(iii\right)$ and $\left(iv\right)\Rightarrow -\frac{1}{3}\le d\le \frac{1}{4}$

Minimum value of $d=-\frac{1}{3}$

Mean $\overline{x}=\frac{6+3d}{4}=\frac{5}{4}$