Question

Suppose that customers arrive at a bank at a rate of 10 per hour. Assume that the number of customer arrivals X follows a Poisson distribution.

Find the probability of more than 25 people arriving within the next two hours using the Poisson mass function.

Find the probability of more than 25 people arriving within the next two hours using the normal approximation to the Poisson.

Compute the percent relative difference between the exact probability computed in part 1 and the normal approximation computed in part 2, i.e., 100 × (Approx−Exact)/Exact %. Comment on the adequacy of the approximation.

please ensure writing is legible

Answer 1

Answer :- Suppose that customers arrive at a bank at a rate of 10 per hour. Assume that the number of customer arrivals X follows a Poisson distribution.

Given that :-

• A customers arrive at bank at a rate of 10 per hour.

[ λ = 10 Per hours ]

1) probability of more than 25 people arriving within the next two hours.

( Using Poisson cumulative table )

=> P(More than 25) = P(X > 25)

=> 1 - P(X < 25) = 1 - P(X < 24)

=> 1 - 0.99998

=> 0.00002

2) probability of more than 25 people arriving within the next two hours.

( using the normal approximation with mean μ = 10)

=> Standard deviation (σ) = √λ

= √10

( σ = 3.1623 )

• For X = 25

=> Z = (X - μ)/σ

= (25 - 10)/ 3.1623

= 15 /3.1623

= 4.7434

=> P(X ≥ 25) = P(Z > 4.7434)

=> 0.0000

3) Here percentage of difference is :-

=> (0.0000 - 0.00002)/ × 100

=> 0.002℅