Please helpok it's blurry sorry
2 answers:
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Answer:
a) I would use the following distribution fir the amount of applicants arriving in a 20 minute period
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
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
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.