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giddy
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This is a problem from my A levels Stats2 book. I understood the problem but one of my answers doesn't seem to be correct according to the book so I thought I better be sure!
A piece of laminated plywood consists of 3 pieces of wood of type A and 2 pieces of type B. The thickness of type A has a mean 2mm and variance 0.04mm2 and The thickness of type B has a mean 1mm and variance 0.01mm2. Find the mean and variance of the thickness of the laminated plywood.
E(aX + bY) = aE(X) + bE(Y)
Var(aX + bY) = a2Var(X) + b2Var(Y)
so T = 3A + 2B
E(T) = 3E(A) + 2E(B) = 8mm (This is correct)
Now : Var(T) = 32Var(A) + 22Var(B)
=9*0.04 + 4*0.01 = 0.36+0.04 = 0.4 (Which is wrong... according to the book the answer should be 0.14mm2)
How? What am I doing wrong?
Homework Statement
A piece of laminated plywood consists of 3 pieces of wood of type A and 2 pieces of type B. The thickness of type A has a mean 2mm and variance 0.04mm2 and The thickness of type B has a mean 1mm and variance 0.01mm2. Find the mean and variance of the thickness of the laminated plywood.
Homework Equations
E(aX + bY) = aE(X) + bE(Y)
Var(aX + bY) = a2Var(X) + b2Var(Y)
The Attempt at a Solution
so T = 3A + 2B
E(T) = 3E(A) + 2E(B) = 8mm (This is correct)
Now : Var(T) = 32Var(A) + 22Var(B)
=9*0.04 + 4*0.01 = 0.36+0.04 = 0.4 (Which is wrong... according to the book the answer should be 0.14mm2)
How? What am I doing wrong?