Question

A new diner opens on our street. It will be open 24 hours a day, seven days a week. It is assumed that the inter-arrival times between customers will be i.i.d.(independent and identically distributed) exponential with mean 10 minutes. Approximate the probability that the 120th customer will arrive after the first 21 hours of operation. Enter your answer to 4 decimal places.

Answer 1

Answer:

0.2912

Step-by-step explanation:

Given that:

Duration new diner will be opened is24 hours a day. It is assumed that the Inter arrival times between customers will be i.i.d exponential with mean 10 minutes

A collection of random variables is independent and identically distributed (i.i.d) if each random variable has the same probability distribution as the others and all are mutually independent.

Expected time for 120 customers = 120 ×10 = 1200 minutes

Standard deviation = $\sqrt{120*10^{2} }$ = 109.5445

P(It takes more than 21 hours) = 21 × 60 =1260 minutes

Probability = P(X > 1260) = P(Z > ((1260-1200)109.5445)

P(X > 1260) = P(Z > (60/109.5445)

P(X > 1260) = P(Z>0.55)

P(X > 1260) = 1 - P(Z<0.55) = 1 - 0.7088 = 0.2912

The probability that the 120th customer will arrive after the first 21 hours of operation is 0.2912