Question 1026509
Let X = random variable for the score
PMF:
X=x 1 2 3 4 5 6
p(X=x) 0 0 0 0.50 0.40 0.10 
==> E(X) = 1*0+2*0+3*0+4*0.50+ 5*0.40 +6*0.10 = 2+2+0.6 = 4.6, the mean of the distribution.

PMF of {{{X^2}}}
{{{X^2 = x^2}}}     1     4     9     16      25     36
p({{{X^2 = x^2}}})  0     0     0    0.50    0.40    0.10 
==> {{{E(X^2) = 1*0+4*0+9*0+16*0.50+ 25*0.40 +36*0.10 = 8+10+3.6 = 21.6 }}},

==>{{{ Var(X) = E(X^2) - (E(X))^2 = 21.6 - 4.6^2 = 0.44}}}