Say X1 is one observation of the 60.E(X1) is 0 and V(X1) is 1/12 (common properties of a uniform dist)Then if S = Sum of all Xs 1 > 60E(S) = E(X1+X2+...+X60) = 60E(X1) = 0V(S) = V(X1 + X2 +...+ X60) = 60V(X1) = 5Since N = 60 you are able to use the normal distribution with mean = 0 and Variance = 5One note is that you need to multiply your resulting probability by 2 as it is a 2tailed test. You need to find (Given that T = true sum value) P( S > T + 3) and P( S < T  3)
