Question

The permeability of a membrane used as a moisture barrier in a biological application depends on...

The permeability of a membrane used as a moisture barrier in a biological application depends on the thickness of three integrated layers. Let X1 denote the thickness of layer 1, X2 denote the thickness of layer 2, and X3 denote the thickness of layer 3. X1, X2, and X3 are normally distributed with means of 1.1, 1.3, and 1.5 millimeters, respectively. The standard deviations are 0.1, 0.3, and 0.5 millimeters, respectively.

a Suppose X1, X2, and X3 are independent. Determine the mean and variance of the total thicknessof the three layers.

b Suppose X1, X2, and X3 are independent. Find the probability that the total thickness of the three layers is less 3.7 millimeters.

c Suppose that the correlation between layers 1 and 2 is 0.6, between layers 1 and 3 is 0.4, and between layers 2 and 3 is 0.3. Determine the mean and variance of the total thickness of the three layers.

Homework Answers

Answer #1

X1 ~ N(1.1, 0.1)

X2 ~ N(1.3,0.3)

X3~N(1.5,0.5)

a) Let the total thickness is denoted by U,

U = X1+X2+X3

MEan = E(U) = E(X1)+E(X2)+E(X3) = 1.1+1.3+1.5 =3.9

V(U) = V(X1)+V(X2)+V(X3) = 0.12 +0.32 + 0.52 = 0.01+0.09+0.25 =0.35

b)

c) It is given that correlation between layers are :

, ,

we know that ,

The mean of the total thickness of the three layers

E(U) = E(X1)+E(X2)+E(X3) = 1.1+1.3+1.5=3.9

The mean of the total thickness of three layers = 3.9

V(U) = V(X1)+V(X2)+V(X3)+2Cov(X1,X2)+2Cov(X1,X3)+2Cov(X2,X3)

V(U) = 0.12 +0.32+0.52 +2*(0.018+0.02+0.045) =0.35+2*0.083 =0.516

The variance of the total thickness of the three layers = 0.516

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