A manufacturer produces items with a rectangular base. Let random variables denote the respective length (X) and width (Y) of the rectangle (in cm)

Question

Suppose X and Y are independent with the following parameter values.

E(X) = 5, Var(X) = 0.19, E(Y) = 6, Var(Y) = 0.35

The perimeter is then defined as 2X + 2Y.

(Round your answers to 4 decimal places, if needed.)

(a) What is the mean of the perimeter of the rectangle?

(b) What is the standard deviation of the perimeter of the rectangle?

Answer 1

E(X) = 5

E(Y) = 6

perimeter = 2X+2Y

a) perimeter mean = 2(5) + 2(6) = 10+12 = 22 cm

b) VarX= .19, VarY =.35

Var2X +Var2Y = 2^2VarX + 2^2VarY = 4VarX +4VarY+2Cov(2X,2Y)

= 4(.19)+4(.35) +2Cov(2X,2Y)

= .76+1.4 +2Cov(2X,2Y)

= 2.16 + 2Cov(2X,2Y)

Cov(2X,2Y) = (1/2)Var(2X+2Y) - VarX-VarY

= (1/2)(Var2X +Var2Y) -VarX-VarY

=(1/2)(4VarX +4VarY) - VarX-VarY

= (1/2(2.16)- .19-.35

= 1.08-.19-.35

= 1.08-.54

= .54

2.16 +2(.54)

= 2.16 +1.08

= 3.24 = Variance of the Perimeter

standard deviation = square root of the Variance = sqr(Var)

= sqr3.24 = 1.8

(No guarantees of error free arithmetic calculations above, but if someone else comes up with the same answer, odds are good it's right. if someone has a different answer, odds are 100% certain one of us is wrong. Main point of the above is to calculate the Covariance, which was not expressly given.)