Statistic Question

Question

Ali lives in a small town where there is also an army base. There is a single ATM in this town. Suppose that Ali approaches this ATM to withdraw some cash, and sees three people queued up before him. The one at the front is just starting his transaction. Ali knows that the soldiers from the near-by base are extremely disciplined, and whenever they use the ATM, they do one of the three transaction with transaction type 1 taking exactly 30 seconds, type 2 taking exactly 1 minute, and type 3 taking exactly 3 minutes. From previous encounters, Ali estimates that transactions of type 1 are done 30% of the time, transactions of type 2 are done 50% of the time, and transactions of type 3 are done 20% of the time, by the soldiers. The civilians are not as disciplined as the soldiers and the amount of time they spend at the ATM (regardless of the transaction type) has the following cdf: [0, |1-e-2(z-1), x > 1, x < 1, Fc(x) = where the random variable C is in minutes. The first two people have their uniforms on, so it is certain that they are soldiers. But the one in front of Ali has no uniform on, so Ali cannot be certain whether he's a soldier or not. Taking into account the population of the town, the number of soldiers in the base, the demeanor of the person etc., Ali estimates the probability that the person in front of him is a soldier at 0.7. (a) Find the distribution of the time Ali waits at the ATM before his turn. (b) Compute the probability that Ali waits more than 5 minutes.

Answer 1

We have a mixture of a multinomial distribution with n = 3 (70%)and a multinomial distribution with n = 2 combined with a negative exponential distribution. (30%)

Let x1 = the number of type 1s, x2 = the number of type 2s, and x3 = the number of type 3s.

px1 = .3, px2 = .5, px3 = .2

For the multinomial with n = 3, multiplying by the probability that n = 3 is .7, we get

P(x1, x2, x3) = .7*3!/x1!x2!x3! .3^x1 .5^x2 .2^x3 and T = .5x1 + x2 + 3x3

P(1.5) = 0.0189

P(2) = 0.0945

P(2.5) = 0.1575

P(3) = 0.0875

P(4) = 0.0378

P(4.5) = 0.126

P(5) = 0.105

P(6.5) = 0.0252

P(7) = 0.042

P(9) = 0.0056

When we have the non-military person as the third person, we calculate the multinomial probability with n=2 and multiply by .3 = .3*2!/x1!x2!x3! .3^x1 .5^x2 .2^x3, then get the beginning time of T = .5x1 + x2 + 3x3 + 1

followed by the negative exponential distribution.

0, 0.027 - 0.027e^-2(t - 2) for t < 2, t > 2

0, 0.09 - 0.09e^-2(t - 2.5) for t < 2.5, t > 2.5

0, 0.075 - 0.075e^-2(t - 3) for t < 3, t > 3

0, 0.036 - 0.036e^-2(t - 4.5) for t < 4.5, t > 4.5

0, 0.06 - 0.06e^-2(t - 5) for t < 5, t > 5

0, 0.012 - 0.012e^-2(t - 7) for t < 7, t > 7

Thus, our distribution function is:

0 for t < 1.5

.0189 for 1.5 <= t < 2

.1404 - .027e^(4-2t) for 2 <= t <2.5

.3879 - .027e^(4-2t) -.09e^(5–2t) for 2.5 <= t < 3

.5504 - .027e^(4-2t) -.09e^(5–2t) -.075e^(6–2t) for 3 <= t < 4

.5504 - .027e^(4-2t) -.09e^(5–2t) -.075e^(6–2t) for 3 <= t < 4

.5882 - .027e^(4-2t) -.09e^(5–2t) -.075e^(6–2t) for 4 <= t < 4.5

.7502 - .027e^(4-2t) -.09e^(5–2t) -.075e^(6–2t) - -.036e^(9–2t) for 4.5 <= t < 5

.9152 - .027e^(4-2t) -.09e^(5–2t) -.075e^(6–2t) -.036e^(9–2t) - .06e^(10–2t) for 5 <= t < 6.5

.9404 - .027e^(4-2t) -.09e^(5–2t) -.075e^(6–2t) -.036e^(9–2t) - .06e^(10–2t) for 6.5 <= t < 7

.9944 - .027e^(4-2t) -.09e^(5–2t) -.075e^(6–2t) - .036e^(9–2t) - .06e^(10–2t) - .012e^(14–2t) for 7 <= t < 9

1 - .027e^(4-2t) -.09e^(5–2t) -.075e^(6–2t) - .036e^(9–2t) - .06e^(10–2t) - .012e^(14–2t) for 9 <= t

Note that I started the negative exponential distribution at the left endpoint. That saved us on the number of lines written above and did not change the value, as the contribution of its component at the left endpoint is 0

Also, note that the limit of all of the negative exponentials as t goes to infinity equals 1, which gives us the probability that we expect.

b) P(t > 5) = 1 – P(t <= 5)

P(t <= 5) =.9152 - .027e^(4-2t) -.09e^(5–2t) -.075e^(6–2t) -.036e^(9–2t) - .06e^(10–2t) for t = 5 = .8399

1 - .8399 = .1601