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).
x1 0 1 2 µ = 1.2, s2 = 0.56
p(x1) 0.2 0.4 0.4
(a) Determine the pmf of To = X1 + X2. to 0 1 2 3 4 p(to)
(b) Calculate µTo. µTo = How does it relate to µ, the population mean? µTo = · µ
(c) Calculate sTo2. sTo2 How does it relate to s2, the population variance?
sTo2 = · s2
(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 Xi's, what now are the values of E(To) and V(To)?
E(To) =
V(To) =
(e) Referring back to (d), what are the values of
P(To = 8) and P(To = 7)
[Hint: Don't even think of listing all possible outcomes!] (Enter your answers to four decimal places.)
P(To = 8) =
P(To = 7) =

Respuesta :

A commuter passes through two traffic lights on their way to and from work. Let X1 represent the number of lights the commuter must stop at while traveling to work, and X2 represent the number of lights he must stop at. stop when returning from work. It is connected to the population mean.

a) P(To=0) = 0.2*0.2 = 0.04

P(To=1) = 0.2*0.4 + 0.4*0.2 = 0.24

P(To=2) = 0.4*0.4 + 0.2*0.2 + 0.4*0.2 = 0.36

P(To=3) = 0.2*0.4 + 0.4*0.4 = 0.24

P(To=4) = 0.2*0.2 = 0.04

b) µTo = 0*0.04 + 1*0.24 + 2*0.36 + 3*0.24 + 4*0.04 = 2

It is equal to the population mean µ.

c) sTo2 = (0-2)2*0.04 + (1-2)2*0.24 + (2-2)2*0.36 + (3-2)2*0.24 + (4-2)2*0.04 = 0.56

It is equal to the population variance s2.

d) E(To) = 0*0.04 + 1*0.24 + 2*0.36 + 3*0.24 + 4*0.04 + 5*0 + 6*0 + 7*0 + 8*0 = 2

V(To) = (0-2)2*0.04 + (1-2)2*0.24 + (2-2)2*0.36 + (3-2)2*0.24 + (4-2)2*0.04 + (5-2)2*0 + (6-2)2*0 + (7-2)2*0 + (8-2)2*0 = 0.56

e) P(To = 8) = 0

P(To = 7) = 0.2*0.2*0.2*0.2 = 0.0016

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