Solve by poisson distribution: A store has an average of 90 customers per hour. a) What is prob. that for 2 minutes the store does not receive any customers? b) Among 100 identical stores, how many would not receive any customers?

a) What is the probability that for 2 minutes the store does not receive any customers? First, we need to find the average rate of customer arrivals for 2 minutes. There are 60 minutes in an hour, so for 2 minutes, the rate can be calculated as follows: Rate for 2 minutes = (2/60) * 90 customers = 3 customers Now, we can use the Poisson distribution formula to calculate the probability of receiving 0 customers in 2 minutes: P(X = 0) = (e^(-λ) * λ^0) / 0! Where: λ (lambda) is the average rate of customer arrivals for 2 minutes, which is 3 in this case. e is the base of the natural logarithm (approximately 2.71828). P(X = 0) = (e^(-3) * 3^0) / 0! P(X = 0) = (e^(-3) * 1) / 1 P(X = 0) = e^(-3) Using a calculator: P(X = 0) ≈ 0.0498 (rounded to four decimal places) So, the probability that the store does not receive any customers in 2 minutes is approximately 0.0498 or 4.98%.