Question

The number of customers X that visit a diner during lunch is a (μ=6.5)-Poisson random variable.

List the four most likely number of customers to visit and corresponding probabilities in order of likelihood.

a)Probability of most likely number of customers

b)Probability of 2nd most likely number of customers

c)Probability of 3rd most likely number of customer

d)Probability of 4th most likely number of customer

Answer 1

X is the number of customers that visit a diner. X follows a Poisson distribution with mean 6.5.

The Poisson distribution is a uni-modal distribution.

X has the following pdf

where,

and

f(0) = 0.0015

f(1) = 0.0097

f(2) = 0.0317

f(3) = 0.0688

f(4) = 0.1118

f(5) = 0.1453

f(6) = 0.1574

f(7) = 0.1462

f(8) = 0.1188

f(9) = 0.0858

We see that beyond 6 the value of f(x) starts to decrease. Therefore we can say since Poisson is a uni-modal distribution that 6 is the most likely number of customers in order of likelihood and its corresponding probability is 0.1574

a)Most likely value is

x=6 , f(x)=0.1574

b)Second most likely value is

x=7 , f(x)=0.1462

c)Third most likely value is

x=5 , f(x)=0.1453

d)Fourth most likely value is

x=8, f(x)=0.1188