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). x1 0 1 2 μ = 1.3, σ2 = 0.81 p(x1) 0.3 0.1 0.6 (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 σTo2. σTo2 = How does it relate to σ2, the population variance? σTo2 = · σ2 (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) =

Answer 1

(a)

Following table shows all the possible samples and corresponding probabilities:

X1	X2	P(X1=x1)	P(X2=x2)	T0	P(T0=t0)=P(X1=x1)P(X2=x2)
0	0	0.3	0.3	0	0.09
0	1	0.3	0.1	1	0.03
0	2	0.3	0.6	2	0.18
1	0	0.1	0.3	1	0.03
1	1	0.1	0.1	2	0.01
1	2	0.1	0.6	3	0.06
2	0	0.6	0.3	2	0.18
2	1	0.6	0.1	3	0.06
2	2	0.6	0.6	4	0.36
Total					1

Following table shows the probability distribution of T0:

T0	P(T0=t0)
0	0.09
1	0.06
2	0.37
3	0.12
4	0.36
Total	1

(b)

Following table shows the calculations for mean of T0:

T0	P(T0=t0)	T0*P(T0=t0)
0	0.09	0
1	0.06	0.06
2	0.37	0.74
3	0.12	0.36
4	0.36	1.44
Total	1	2.6

So

That is

(c)

Following table shows the calculations for variances:

T0	P(T0=t0)	T0*P(T0=t0)	T0^2*P(T0=t0)
0	0.09	0	0
1	0.06	0.06	0.06
2	0.37	0.74	1.48
3	0.12	0.36	1.08
4	0.36	1.44	5.76
Total	1	2.6	8.38

So

So

(d)

Now T0 is sum of four Xi so from the results of b and c we have

(e)

T0 can be 8 if all the four variables X1, X2, X3 and X4 are 2 so required probability is

P(T0=8) = 0.6*0.6*0.6*0.6= 0.1296

T0 can be seven if three of X1, X2, X3 and X4 are 2 and one is 1. So

P(T0 = 7) = C(4,3) * [ 0.1*0.6*0.6*0.6] = 0.0216

So required probability is

P(T0 >= 7) = P(T0 = 7)+ P(T0 = 8)=0.0216+0.1296 = 0.1512