46. Weightlifting. The Atlas BodyBuilding Company(ABC)

Weightlifting. The Atlas BodyBuilding Company(ABC) sells starter sets of barbells that consistof one bar, two 20-pound weights, and four 5-pound weights. The bars weigh an average of 10pounds with a standard deviation of 0.25 pounds. The weights average the specified amounts, butthe standard deviations are 0.2 pounds for the 20-pounders and 0.1 pounds for the 5-pounders.We can assume that all the weights are normally distributed.a) ABC ships these starter sets to customers in two boxes: The bar goes in one box and the sixweights go in another. What's the probability that the total weight in that second box exceeds60.5 pounds. Define your variables clearly and state any assumptions you make.b) It costs ABC $0.40 per pound to ship the box containing the weights. Because it's anodd-shaped package,though, shipping the bar costs $0.50 a pound plus a $6.00 surcharge. Findthe mean and standard deviation of the company's total cost for shipping a starter set.c) Suppose a customer puts a 20-pound weight at one end of the bar and the four 5-poundweights at the other end. Although he expects the two ends to weigh the same, they might differslightly. What's the probability the difference is more than a quarter of a pound.AnswerStep 1 of 3(a)We can represents the bar with X, 20-pound weight with Y and 5-pound weight with Z.We have given,\n E(X) = 10, E(Y ) = 20, and E(Z) = 5 = 0.25, = 0.2, and = 0.1 X Y Z V ar(X) = 2= 0.25 = 0.0625, = 0.2, and = 0.1 X Y Z 2 2 V ar(Y ) = Y = 0.2 = 0.04 2 2 V ar(Z) = Z = 0.1 = 0.01Let W represent total weight in that second box.We need to find P(W > 60.5 pounds).We can write the total weight W as, W = Y + Y + Z + Z + Z + Z 1 2 1 2 3 4 W = E(W) = E(Y + Y1+ Z +2Z + 1 + Z2) 3 4 W = E(W) = E(Y ) +1E(Y ) + 2(Z ) + E1Z ) + E(2 ) + E(Z3) 4 E(Y ) = E(Y ) 1 E(Y ) =220 E(Z ) = E(Z ) = E(Z ) = E(Z ) = E(Z ) = 5 1 2 3 4\n W = E(W) = 2E(Y ) + 4E(Z ) W = E(W) = 2 × 20 + 4 × 5 = 60 pounds W = Y 1Y 2Z 1Z 2Z 3Z 4 = ar(Y + 1 + Z 2 Z + 1 + Z2) 3 4 W = Y +Y +Z +Z +Z +Z = V ar(Y 1 + V ar(Y ) 2 V ar(Z ) +1V ar(Z ) + 2 ar(Z ) + 3 ar(Z ) 4 1 2 1 2 3 4 = = 2V ar(Y ) + 4V ar(Z ) W Y 1Y 2Z 1Z 2Z 3Z 4 = = 2 × 0.04 + 4 × 0.01 0.3464 W Y 1Y 2Z 1Z 2Z 3Z 4Since all the weights are normally distributed hence thescore, W Z = W Wz score 60.5 pounds corresponding to the value of is, Z = 60.560 1.44 0.3464 P(W > 60.5 pounds) = 1 P(W 60.5 pounds) P(W 60.5 pounds) = P(Z 1.44) = 0.9251(Using the normal probability curve or Z table)\n P(W > 60.5 pounds) = 1 0.9251 = 0.0749 Hence the probability that the total weight in that second box exceeds 60.5 pounds is0.0749.