Question

1. If the number of arrivals at a cell phone kiosk has a rate of 0.80 customers per hour.

to. What is the probability that less than 2 clients arrive in the next half hour? (10 points)

b. What is the probability that in the next three hours between 2 and 4 clients will arrive inclusive? (10 points)

c. Last customer arrived at 2:00 pm, what is the probability that the next customer takes more than 60 minutes to arrive? (10 points)

d. What is the average time between arrivals in minutes? (5 points)

and. What is the medina of the time between arrivals? (5 points)

F. What is the 80% percentile? (5 points)

Answer 1

1.

Let X be the number of customers arrive in t hour.

Average number of customers arrive in t hour = 0.8t

X ~ Poisson( = 0.8t)

For t = 0.5

X ~ Poisson( = 0.4)

Probability that less than 2 clients arrive in the next half hour = P(X < 2)

= P(X = 0) + P(X = 1)

= exp(-0.4) * 0.4^0 / 0! + exp(-0.4) * 0.4^1 / 1!

= 0.670320 + 0.268128

= 0.938448

b)

For t = 3

X ~ Poisson( = 2.4)

Probability that in the next three hours between 2 and 4 clients will arrive inclusive

= P(2 X 4)

= P(X = 2) + P(X = 3) + P(X = 4)

= exp(-2.4) * 2.4^0 / 0! + exp(-2.4) * 2.4^1 / 1! + exp(-2.4) * 2.4^2 / 2!

= 0.2612677 + 0.2090142 + 0.1254085

= 0.5957

c)

For t = 1

X ~ Poisson( = 0.8)

Probability that the next customer takes more than 60 minutes to arrive = Probability that no customers arrive in next 1 hour = P(X = 0)

= exp(-0.8) * 0.8^0 / 0!

= 0.4493

d)

For t = 1 min = (1/60) hour

X ~ Poisson( = 0.8/60)

We know that the time between Poisson arrivals follow exponential distribution with rate = 0.8/60 per minute.

Average time between arrivals in minutes = 1/ = 60/0.8 = 75 minutes

e)

Median of the time between arrivals = ln(2)/ = 0.6931 * 60/0.8 = 51.98 minutes

f)

80% percentile is,

0.8 = 1 - exp(-x)

exp(-0.8x/60) = 0.2

-0.8x/60 = log(0.2)

x = 1.609438 * 60 / 0.8 =  120.71 minutes