X	0	1	2	3
$$P(x)$$	$${{1 - d} \over 4}$$	$${{1 + 2d} \over 4}$$	$${{1 - 4d} \over 4}$$	$${{1 + 3d} \over 4}$$

We know, $$0 \le P(x) \le 1$$

$$\therefore$$ $$0 \le {{1 - d} \over 4} \le 1$$

$$ \Rightarrow 0 \le 1 - d \le 4$$

$$ \Rightarrow - 1 \le - d \le 3$$

$$ \Rightarrow 1 \ge d \ge - 3$$

Also,

$$0 \le {{1 + 2d} \over 4} \le 1$$

$$ \Rightarrow 0 \le 1 + 2d \le 4$$

$$ \Rightarrow - 1 \le 2d \le 3$$

$$ \Rightarrow - {1 \over 2} \le d \le {3 \over 2}$$

Also,

$$0 \le {{1 - 4d} \over 4} \le 1$$

$$ \Rightarrow 0 \le 1 - 4d \le 4$$

$$ \Rightarrow - 1 \le - 4d \le 3$$

$$ \Rightarrow 1 \ge 4d \ge - 3$$

$$ \Rightarrow {1 \over 4} \ge d \ge - {3 \over 4}$$

And,

$$0 \le {{1 + 3d} \over 4} \le 1$$

$$ \Rightarrow 0 \le 1 + 3d \le 4$$

$$ \Rightarrow - 1 \le 3d \le 3$$

$$ \Rightarrow - {1 \over 3} \le d \le 1$$

Common range is $$ = - {1 \over 3}$$ to $${1 \over 4}$$

$$\therefore$$ $$d\, \in \left[ { - {1 \over 3},{1 \over 4}} \right]$$

$$\therefore$$ Minimum value of $$d = - {1 \over 3}$$

We know, mean

$$E(x) = \sum {x\,.\,P(x)} $$

$$ = 0 \times {{1 - d} \over 4} + 1 \times {{1 + 2d} \over 4} + 2 \times {{1 - 4d} \over 4} + 3 \times {{1 + 3d} \over 4}$$

$$ = {{1 + 2d + 2 - 8d + 3 + 9d} \over 4}$$

$$ = {{6 + 3d} \over 4}$$

For $$d = - {1 \over 3}$$, $$E(x) = {{6 + 3 \times - {1 \over 3}} \over 4} = {5 \over 4}$$

$$\therefore$$ $$60E(x) = 60 \times {5 \over 4} = 75$$