2 points DevoreStat9 5..038 There are two traffic lights on a commuter\'s route

Question

2 points DevoreStat9 5..038 There are two traffic lights on a commuter's route to and from work. Let be the number of lights at which the commuter must stop on his way to work, and X2 be the number or lights at which he must stop when retuming from work. Suppose that these two variables are independent, each with the pmf given in the accompanying table (so XI, X, is a random sample of size n-2). My Notes -1.6, 2-0.44 pxi) 0.1 0.2 0.7 (a) Determine the pmf of To = X1 + X2. prto) (b) Calculate ar How does it relate to ' the population mean? c) Calculate 7c2 Haw does it relate to o2, the population variance? (d) Let X3 and X4 be the number of lights at which a stop is required when driving to and from work on a second day assumed independent of the first day. With Tothe sum of all four Xs, what now are the values of ETo) and To)? E(To) = (e) Referring back to (d), what are the values af P(T = a and P T 7) Hint: Don't even think of listing all possible outcomes!] Enter your answers to four decimal places.) Need Help?RedTalik to a Tutor

Explanation / Answer

a) P(t0 = 0) = P(X1 = 0 and X2 = 0) = 0.1 * 0.1 = 0.01

P(t0 = 1) = P(X1 = 1 and X2 = 0) + P(X1 = 0 and X2 = 1) = 0.2 * 0.1 + 0.1 * 0.2 = 0.04

P(t0 = 2) = P(X1 = 1 and X2 = 1) + P(X1 = 2 and X2 = 0) + P(X1 = 0 and X2 = 2)= 0.2 * 0.2 + 0.7 * 0.1 + 0.1 * 0.7 = 0.18

P(t0 = 3) = P(X1 = 1 and X2 = 2) + P(X1 = 2 and X2 = 1) = 0.2 * 0.7 + 0.7 * 0.2 = 0.28

P(t0 = 4) = P(X1 = 2 and X2 = 2) = 0.7 * 0.7 = 0.49

b) t0 = E(t0) = 0 * 0.01 + 1 * 0.04 + 2 * 0.18 + 3 * 0.28 + 4 * 0.49 = 3.2

t0 = 2 *

c) E(t02) = 02 * 0.01 + 12 * 0.04 + 22 * 0.18 + 32 * 0.28 + 42 * 0.49 = 11.12

t02 = E(t02) - (E(t0))2 = 11.12 - 3.22 = 0.88

t02 = 2 * 2

d) E(t0) = 4 * = 4 * 1.6 = 6.4

V(t0) = 4 * 2 = 4 * 0.44 = 1.76

e) P(t0 = 8) = P(X1 = 2 and X2 = 2 and X3 = 2 and X4 = 2) = 0.74 = 0.2401

P(t0 = 8) = P(X1 = 1 and X2 = 2 and X3 = 2 and X4 = 2) + P(X1 = 2 and X2 = 1 and X3 = 2 and X4 = 2) + P(X1 = 2 and X2 = 2 and X3 = 1 and X4 = 2) + P(X1 = 2 and X2 = 2 and X3 = 2 and X4 = 1)

              = 4 * 0.2 * 0.73 = 0.2744

P(t0> 7) = P(t0 = 7) + P(t0 = 8)

              = 0.2744 + 0.2401

              = 0.5145

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