2. -12 points DevoreStat9 3.E.030. My Notes Ask Your An individual who has autom
ID: 3046795 • Letter: 2
Question
2. -12 points DevoreStat9 3.E.030. My Notes Ask Your An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf or Y is the following. p() 0.60 0.25 0.10 0.05 (a) Compute E(Y) E(Y) = (b) Suppose an individual with violations incurs a surcharge of $100y2. Calculate the expected amount or the surcharge. Need Help? .nd..i IT-tonteri 3. -/4 points DevoreStat9 3.E.039. My Notes Ask Your A chemical supply company currently has in stock 100 lb of a certain chemical, which it sells to customers in 7-lb batches. Let X-the number of batches ordered by a randomly chosen customer, and suppose that X has the following pmt 1 2 3 4 p(x) 0.2 0.40.3 01 Compute E(X) and Vx) E(X) v(x) = batches batches2 Compute the expected number of pounds left after the next customer's order is shipped and the varlance of the number of pounds left. [Hint: The number of pounds left is a linear function of X.] expected weight left variance of weight left lb Ib2 Need Help?Readit Talk to a TutorExplanation / Answer
2) a) E(Y) = 0 * 0.6 + 1 * 0.25 + 2 * 0.1 + 3 * 0.05 = 0.6
b) E(Y2) = 02 * 0.6 + 12 * 0.25 + 22 * 0.1 + 32 * 0.05 = 1.1
E(100Y2) = 100 * 1.1 = 110
3)
a) E(X) = 1 * 0.2 + 2 * 0.4 + 3 * 0.3 + 4 * 0.1 = 2.3 batches
E(X2) = 12 * 0.2 + 22 * 0.4 + 32 * 0.3 + 42 * 0.1 = 6.1
Var(X) = E(X2) - (E(X))2 = 6.1 - 2.32 = 0.81 batches2
b) weight left = 100 - 7 * X
E(weight left) = E(100 - 7X) = 100 - 7 * E(X) = 100 - 7 * 2.3 = 83.9 lb
V(weight left) = V(100 - 7X) = (-7)2 * V(X) = 49 * 0.81 = 39.69 lb2
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