A-Level Normal distribution: worked solution

Question

A deluxe seesaw uses 2 cylindrical blocks, 3 cuboidal blocks, 7 screws and 1 plank. Drilling removes 5% of each cylindrical block's mass and 3% of each cuboidal block's mass. Screw mass ~ N(8,1.2²) and plank mass ~ N(540,8²), all chosen at random. Find the probability that the finished model's total mass is above 1785 g.

From the earlier parts of this question the component masses are C ~ N(280, 6²) for a cylindrical block, D ~ N(220, 5²) for a cuboidal block, S ~ N(8, 1.2²) for a screw and P ~ N(540, 8²) for the plank, all independent. Drilling removes 5% of a cylindrical block, so 95% is left and a drilled cylinder has mass 0.95C. Drilling removes 3% of a cuboidal block, so 97% is left and a drilled cuboid has mass 0.97D. The finished model is the sum of 2 + 3 + 7 + 1 = 13 separate, independently chosen items: T = 0.95C1 + 0.95C2 + 0.97D1 + 0.97D2 + 0.97D3 + S1 + S2 + ... + S7 + P. Mean: expectation is linear, so multiply each mean by its scale factor and add. E(T) = 2(0.95)(280) + 3(0.97)(220) + 7(8) + 540 = 2(266) + 3(213.4) + 56 + 540 = 532 + 640.2 + 56 + 540 = 1768.2. Variance: multiplying a variable by a constant a multiplies its variance by a², and variances of independent variables add. Var(T) = 2(0.95²)(6²) + 3(0.97²)(5²) + 7(1.2²) + 8² = 2(0.9025)(36) + 3(0.9409)(25) + 7(1.44) + 64 = 64.98 + 70.5675 + 10.08 + 64 = 209.6275. (Note this is 2 x 0.95² x 36 and NOT (2 x 0.95)² x 36: two blocks chosen at random are two independent variables added, not one variable doubled.) A sum of independent normal variables is normal, so T ~ N(1768.2, 209.6275), with standard deviation sqrt(209.6275) = 14.4785. Standardise: P(T > 1785) = P(Z > (1785 - 1768.2)/14.4785) = P(Z > 1.1603) = 1 - 0.8770 = 0.1230. So the probability the finished model weighs more than 1785 g is 0.123 (3 s.f.).