Question

Every day, patients arrive at the dentist’s office. If the Poisson distribution were applied to this process:

a.) What would be an appropriate random variable? What would be the exponential-distribution counterpart to the random variable?

b.)If the random discrete variable is Poisson distributed with λ = 10 patients per hour, and the corresponding exponential distribution has x = minutes until the next arrival, identify the mean of x and determine the following:

1. P(x less than or equal to 6) 2. P(x greater than or equal to 7.5)

c.)The timing of arrival of the patients is found to be exponentially distributed with mean time between arrivals of 6 minutes. This Wednesday, the dentist must leave for 12 minutes during her office hours. What is the probability that a patient will arrive while the doctor is out?

d.)What is the probability that a patient will arrive between 6 and 12 minutes from now?

Answer 1

(a) Here the appropriate random variable is the number of patients arrive at the dentist office. Here exponetial distribution counterpart is the time taken by a patient at the dentist' office.

(b)  Poisson distributed with λ = 10 patients per hour

x = minutes until the next arrival

Here x = 1/10 hour = 6 mins

(c) P(x <= 6) = EXPONDIST(6; 6; true) = 0.6321

Expected number of patients to be arrived in 12 minutes = 12/6 = 2

P(A patient will arrive in next 12 minutes) = 1 - P(No arrive) = 1- e-2 = 0.8647

(d) P(A patient will arrive between 6 and 12 minutes) = 1 - P(No patient will arrive in between 6 and 12 minutes)

= 1- e-1 = 0.6321