Complete the probabilistic estimating problem (below) from the class notes on pr

Question

Complete the probabilistic estimating problem (below) from the class notes on probabilistic moments. A construction project requires: 10,000 cubic yards of earth excavation, E[E] = $3.00/yd^3, V_E = 10%. 1,000 cubic yards of rock excavation, E[R] = $20.00/yd3, VR = 25% 5,000 cubic yards of excavated soil can be sold for a credit of E[S] = -$1.00/yd^3, V_S = 5% For the total cost, C = E + R + S (where S is negative), find E[C], var_c, sigma_c, V_c. The percent contribution to the total uncertainty for each variable is the a^2var_x term divided by the total variance. Find the contribution of each variable. From problem 1, assuming C is normally distributed, find how much money the owner should have available to be 95% sure there is sufficient funding. The travel time to the Detroit airport in the morning and afternoon can be expressed as the following normally distributed random variables: E[M] = 120 minutes, V_M= 20% E[A] = 90 minutes, V_M = 10% For each variable, find the probability that a trip takes more than 100 minutes. Joe travels in the morning, and Sue travels in the afternoon. What is the probability that both make it in under 100 minutes? From homework 3, the unconfined compression strength of clay, q_u, can be assumed to be a random variable with E[q_u] = 1300 psf and sigma_qu = 306 psf. Assuming that the strength is normally distributed, what is the probability that a random sample has a strength between 1400 and 1800 psf?

Explanation / Answer

Answer to question# 1)

E = 3 P1 = 0.10

R = 20 P2=0.25

S = 1 P3=0.05

.

E(C) = E*P1 + R*P2 + S*P3

E(C ) = 3 * 0.10 + 20 * 0.25 + 1 *0.05

E(C ) = 0.30 + 5 + 0.05 = 5.35

E(C ) = 5.35

.

VAR(C ) = E[C^2] - (E[C])^2

We got E[C] = 5.35

Now let us find E[C^2]

E[C^2] = E^2*P1 + R^2*P2 + S^2*P3

E[C^2] = 3^2 *0.1 + 20^2 * 0.25 + 1^2 * 0.05

E[C^2] = 0.9 + 100 + 0.05

E[C^2] = 100.95

.

Thus Var[C] = 100.95 - 5.35^2

Var[C] = 72.3275

.

(C) = Square root of [72.3275]

(C) = 8.5046

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