Question 1180232
(a)  
X     p(X=x)
1     6/23
2     3/23
3     2/23
4     3/23
5     6/23


(b)  {{{E(X) = 1*(6/23) + 2*(3/23) + 3*(2/23) + 4*(3/23) + 5*(6/23) = 60/23}}}


(c) {{{E(X^2) =  1^2*(6/23) + 2^2*(3/23) + 3^2*(2/23) + 4^2*(3/23) + 5^2*(6/23) = 234/23}}}, and in turn this gives

{{{Var(X) = E(X^2) - (E(X))^2 = 234/23 - (60/23)^2 = 1782/529}}}

{{{SD(X) = sqrt(Var(X)) = sqrt(1782/529) = 1.8354}}} to 4 d.p.


(d)  The probability mass function will be {{{p(x) = (2+(x-3)^2)/23}}}, for x = 1, 2, 3, 4, 5.  (This can be seen by observing the symmetry of values around x = 3.)


(e) {{{P(X = 4,5) = 3/23 + 6/23 = 9/23}}}