I can\'t find the answer of (d) and (e). Please, explain it more details. Thanks

Question

I can't find the answer of (d) and (e). Please, explain it more details. Thanks.

If two loads are applied to a cantilever beam as shown in the accompanying drawing, the bending moment at 0 due to the loads is a_1X_1 + a_2X_2. Suppose that X_1 and X_2 are independent rv's with means 4 and 8 kips, respectively, and standard deviations 1.4 and 2.8 kip, respectively. If a_1 = 9 ft and a_2 = 18 ft, what is the expected bending moment and what is the standard deviation of the bending moment? (Round your standard deviation to three decimal places. expected bending moment 180 klp-ft standard deviation 51.95 kip-ft If X_1 and X_2 are normally distributed, what is the probability that the bending moment will exceed 75 kip-ft? (Round your answer to four decimal places.) 0.9783 Suppose the positions of the two loads are random variables. Denoting them by A_1 and A_2 assume that these variables have means of 9 and 18 ft, respectively, that each has a standard deviation of 0.5, and that all A_j's and X_i's are Independent of one another. What Is the expected moment now? 180 kip-ft For the situation of part (c), what is the variance of the bending moment? 2732.29 kip-ft^2 If the situation is as described in part (a) except that Corn(X_1, X_2) = 0.5 (so that the two loads are not independent what is the variance of the bending moment? 2860.92 kip-ft^2

Explanation / Answer

let X=A1X1+A2X2

(c)expected moment=E(X)=E(A1X1+A2X2)=A1E(X1)+A2E(X2)=4*9+8*18=180

(d)Variance(X)=Variance(A1X1+A2X2)=A1*A1*Varaince(X1)+A2*A2*Varaince(X)

=9*9*0.5*0.5+18*18*0.5*0.5=101.25

variance is square of standard deviation (SD)

given that SD(X1)=0.5 and SD(X2)=0.5

(e) let X=a1X1+a2X2

var(X)=var(a1X1+a2X2)=a1*a1*var(X1)+a2*a2*var(X2)+2*a1*a2*cov(X1,X2)

=a1*a1*var(X1)+a2*a2*var(X2)+2*a1*a2*Corr(X1,X2)*SD(X1)*SD(X2)

=9*9*1.4*1.4+18*18*2.8*2.8+2*9*18*0.5*1.4*2.8=3333.96

since corr(X1,X2)=Cov(X1,X2)/(SD(X1)*SD(X2))

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