**6. Correct. The answer is false.**

** **

a. Let X and Y be the total number of tons produced each day by the two
machines. Then:

E(X) = E(Y) = 1(1/6) + 2(1/6) + . . .+ 6(1/6) = 7/2 and

E(X^{2}) = E(Y^{2}) = 1(1/6) + 2^{2}(1/6) + . . .+ 6^{2}(1/6)
= 91/6.

Remember that Var[X] = E[X^{2}] - (E[X])^{2}.^{ }Then, Var (X)=Var(Y) = 91/6 - (7/2)^{2}
= 35/12.^{}

Since X and Y are independent, Var(X+Y) = Var (X) + Var(Y) =
35/6.

b. The standard deviation is the square root of the variance Thus: s_{X+Y} = (35/6)^{1/2}