There are two traffic lights on a commuter\'s route to and from work. Let X1 be

Question

There are two traffic lights on a commuter's route to and from work. Let X1 be the number of lights at which the commuter must stop on his way to work, and X2 be the number of lights at which he must stop when returning from work. Suppose that these two variables are independent, each with the pmf given in the accompanying table (so X1, X2 is a random sample of size n-2) 0 -1.5, 2-0.65 p(%) 0.2 0.1 0.7 (a) Determine the pmf of To-X1 + X2 0 2 3 4 p(to) (b) Calculate HTo How does it relate to , the population mean? (c) Calculate How does it relate to . the population variance? 2 (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 To = the sum of all four Xis, what now are the values of E(To) and V(To)? E(To) = (e) Referring back to (d), what are the values of P(70-8) and PT, places.) 7) [Hint: Don't even think of listing all possible outcomes!] (Enter your answers to four decimal

Explanation / Answer

(a) Here t0 = X1 + X2

t0 = 0 when X1 = 0 and X2 = o

P(0) = 0.2 * 0.2 = 0.04

t0 = 1 when X1,X2 = { (1,0), (0,1)}

P(0) = 2 * 0.2 * 0.1 = 0.04

t0 = 2 when X1,X2 = { (2,0), (0,2), (1,1)}

P(0) = 0.2 * 0.7 + 0.2 * 0.7 + 0.1 * 0.1 = 0.29

t0 = 3 when X1,X2 = { (2,1), (1,2)}

P(0) = 0.1 * 0.7 + 0.1 * 0.7 = 0.14

t0 = 4 when X1,X2 = { (2,2)

P(0) = 0.7 * 0.7 = 0.49

so P(T) = 0.04 for t0 = 0

= 0.04 for t0 =1

= 0.29 for t0 = 2

= 0.14 for t0 = 3

= 0.49 for t0 = 4

(b) T0 = 0 * 0.04 + 1 * 0.04 + 0.29 * 2 + 3 * 0.14 + 4 * 0.49 = 3

T0  =2

(c) 2 To  = 0.04 * 32 + 0.04 * 22 + 0.29 * 12 + 0.14 * 0 + 0.49 * 12 = 1.3

2 To  = 2 2

(d) Here now there are 4 lights so

E(T0 ) = 4 * 1.5 = 6

Var(T0) = 4 * 0.65 = 2.6

(E) P(T0) = 8 will be when all lights will take 2 minutes

P(T0 = 8) = 0.74 = 0.2401

P(T0 >= 7) = P(T0 =7) + P(T0 = 8)

Here, P(T0 = 8)  = 0.2401

P(T0 =7) when there (x1,x2,x3,x4) = (1,2,2,2), (2,1,2,2), (2,2,1,2), (2,2,2,1)

P(T0 =7) = 4 * 0.7 * 0.7 * 0.7 * 0.1 = 0.1372

P(T0 >= 7) = P(T0 =7) + P(T0 = 8)  = 0.1372 + 0.2401 = 0.3773

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