Answer to Question #135941 in Statistics and Probability for Pethias Lungo

Question #135941

The arrival rate of customers arriving at a bank counter follows a Poisson distribution with a mean rate of 4 per 10 minutes interval. Find the probability that i. Exactly 0 customer will arrive in 10 minutes interval ii. Exactly 2 customers will arrive in 10 minutes interval iii. At most 2 customers will arrive in 10 minutes interval iv. At least 3 customers will arrive in 10 minutes interval 


1
Expert's answer
2021-04-14T11:50:32-0400

Let NN be the number of customers arriving in 10 minutes interval.

NN has Poisson(4)Poisson(4) distribution, hence P(N=n)=4ne4n!P(N=n)=\frac{4^ne^{-4}}{n!}.

i. P(N=0)=40e40!=e4P(N=0)=\frac{4^0e^{-4}}{0!}=e^{-4}.

ii. P(N=2)=42e42!=8e4P(N=2)=\frac{4^2e^{-4}}{2!}=8e^{-4}.

iii.

P(N2)=P(N=0)+P(N=1)+P(N=2)=e4+41e41!+8e4=13e4.P(N\leq2)=\\ P(N=0)+P(N=1)+P(N=2)=\\ e^{-4}+\frac{4^1e^{-4}}{1!}+8e^{-4}=13e^{-4}.

iv.

P(N3)=P(N>2)=1P(N2)=113e4.P(N\geq3)=P(N>2)=1-P(N\leq2)=1-13e^{-4}.

Answer: i. e40.0183e^{-4}\approx0.0183, ii. 8e40.14658e^{-4}\approx 0.1465, iii. 10e40.183110e^{-4}\approx 0.1831, iv. 113e40.76191-13e^{-4}\approx 0.7619.


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Comments

Assignment Expert
14.04.21, 18:50

Dear Keshav Goyal, thank you for correcting us.

Keshav Goyal
13.04.21, 13:29

It would be total of 13e(-4) in the third part because 1e(-4) + 4e(-4) + 8e(-4) = 13e(-4) which is equal to 0.2379.

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