Question

If the number of arrivals in a cell phone shop follows a Poisson distribution, with a reason of 10 clients per hour:

What is the probability that in the next half hour, 4 clients arrive?

What is the probability that in the next two hours, between 18 and 22 clients arrive?

What is the average time between arrivals?

What is the median of the time between arrivals?

What is the probability that the time that transpires for the next arrival is between 5 and 10 minutes?

Answer #1

1)expected number of clients in half hour =10*30/60=5

hence probability that in the next half hour, 4 clients
arrive=P(X=4)=e^{-5}*5^{4}/4! =0.1755

2)expected number of clients in 2 hour =2*10=20

probability that in the next two hours, between 18 and 22 clients arrive

=P(18<=X<=22)=(e^{-20}20^{18}/18!+e^{-20}20^{19}/19!+e^{-20}20^{20}/20!+e^{-20}20^{21}/21!+e^{-20}20^{22}/22!)=0.4236

3)average time between arrivals =60/10 =6 minute=

4)

median of the time between arrivals =ln(2)=4.159

5)

probability that the time that transpires for the next arrival is between 5 and 10 minutes=P(5<X<10)

=(1-e^{-10/6})-(1-e^{-5/6})=0.2457

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