Question

2. At the test center giving the exam from the previous problem, the firm needs to plan for proper staffing. Historical data suggests that the average number of applicants arriving per 20-minute period is 1.5. a. What is the best distribution to represent the number of applicant arrivals per 20-minute period? b. If five or more applicants arrive during an hour, the testing center needs to add one more staff member for that hour to proctor and evaluate exams. If the test center operates 50 hours per week, during how many of those 50 hours do we expect to need the extra staff member?

Answer 1

Answer:

a) I would use the following distribution fir the amount of applicants arriving in a 20 minute period

$P_X(k) = \frac{e^{-1.5} \, * \, 1.5^k}{k!}$

b) In a 50 hour period, we will expect to need 0.4679*50 = 23.3948 hours with the extra staff member

Step-by-step explanation:

a) For a total amount of arrivals in a 20 minute period i would use a Poisson distribution with parameter λ = 1.5 (the average). The distribution X is given by this formula

$P_X(k) = \frac{e^{-1.5} \, * \, 1.5^k}{k!}$

b) For one hour, the average will be 1.5*60/20 = 4.5 applicants. The distribution Y for the amount of applicants in one hour is given by the following formula

$P_Y(k) = \frac{e^{-4.5} \, * \, 4.5^k}{k!}$

First, we want to find the probability of Y being greater then or equal to 5. We can obtain the probability of the complementary event and substract it from 1. That event is equal to the probability of Y being equal to 0,1,2,3 or 4

P(Y=0) = e^{-4.5}

P(Y=1) = e^{-4.5}* 4.5

P(Y=2) = e^{-4.5} * 4.5²/2 = e^{-4.5} * 10.125

P(Y=3) = e^{-4.5} * 4.5³/6 = e^{-4.5} * 15.1875

P(Y=4) = e^{-4.5} * 4.5⁴/24 = e{-4.5} * 17.08594

Thus,

P(Y < 5) = P(Y=0)+P(Y=1)+P(Y=2)+P(Y=3)+P(Y=4) = e^{-4.5} * (1+4.5+10.125+15.1875+17.08594) = 0.5321

Therefore,

P(Y ≥ 5) = 1-0.5321 = 0.4679

In a 50 hour period, we will expect to need 0.4679*50 = 23.3948 hours with the extra staff member.