Question

You are waiting at a bus stop and can take any one of two buses Bus 1 or Bus 2. Bus 1 comes every 5 minutes and Bus 2 every 10 minutes. Further assume that the waiting times are memoryless in the sense that the amount of time since the previous bus arrived does not affect how much time to wait until the next bus comes and that the waiting times for each of the three buses are independent. (a) (4 points) What distribution would you use to model how long it it will take for Bus 1 to arrive (specify b) (4 points) Use the distributions you determined in part (a). How long do you expect to wait for either (c) (4 points) What distribution would you use to model the total number of buses 1, and 2 arrive over a d) (4 points) Use the distribution you determined in part (c). What is the chance that 2 or fewer buses all related parameters)? Do the same for Bus 2. of the 2 buses, 1 or 2 to arrive? 30 minute period? (out of any of Bus 1 or 2) arrive over a 30 minute period?

Answer 1

(a)

Since, the waiting times for both the buses are memoryless, the waiting times can be modeled by the exponential distribution.

Let T1, T2 be the waiting times for bus 1 and bus 2 respectively. Then,

T1 ~ Exponential($\lambda$ = 1/5) and T2 ~ Exponential($\lambda$ = 1/10) (Note that the parameter (rate) $\lambda$ = 1/mean)

(b)

We will wait for the time T = min(T1, T2)

CDF of T is,

F(t) = P(T $\le$ t) = 1 - P(T > t) = 1 - P(T1 > t, T2 > t) {Because T = min(T1, T2)}

= 1 - P(T1 > t) * P(T2 > t) {T1 and T2 are independent}

= 1 - exp(-t/5) * exp(-1/10)

= 1 - exp(-3t/10) = 1 - exp(-0.3t)

PDF of T is,

f(t) = F'(t) = 0.3 exp(-0.3t)

Thus, T ~ Exponential($\lambda$ = 0.3)

Expected time for either of the buses to arrive = 1/$\lambda$ = 1/0.3 = 10/3 minutes

= 3.33 minutes

(c)

We know that the number of arrivals with arrival times modeled as exponential distribution follows Poisson distribution.

Expected number of buses 1 to arrive in 30 minute = 30/5 = 6

Expected number of buses 2 to arrive in 30 minute = 30/10 = 3

Let N1 , N2 be the number of buses arrive in 30 minute period. Then,

N1 ~ Poisson($\mu$ = 6) and N2 ~ Poisson($\mu$ = 3)

where $\mu$ is the parameter of the distribution and is equal to the mean number of buses in 30 minute period.

We know that the sum of Poisson random variables with parameter $\mu_1$ and $\mu_2$ is Poisson random variable with parameter $\mu_1 + \mu_2$

Let N be the total number of buses 1 and 2 arrive in 30 minute period. Then,

N = N1 + N2 ~ Poisson($\mu$ = 6 + 3 = 9)

N ~ Poisson(9)

(d)

Chance that 2 or fewer buses arrive over a 30 minute period,

= P(N $\le$ 2) = P(N = 0) + P(N = 1) + P(N = 2)

= exp(-9) * 9⁰ / 0! + exp(-9) * 9¹ / 1! + exp(-9) * 9²​​​​​​​ / 2!

= 0.0001234098 + 0.001110688 + 0.004998097

= 0.006232195