To solve this problem, we first need to find the average rate of arrivals per minute. This is given by the total number of arrivals divided by the total time. In this case, we have 17 customers over 85 minutes, so the average rate λ is 17/85 = 0.2 customers per minute.

The Poisson distribution gives the probability of a given number of events happening in a fixed interval of time, if these events happen with a known average rate and independently of the time since the last event. The formula for the Poisson distribution is:

P(X=k) = λ^k * e^-λ / k!

where:
- P(X=k) is the probability of k events,
- λ is the average rate of events per interval,
- e is the base of the natural logarithm (approximately 2.71828),
- k! is the factorial of k.

We want to find the probability that more than 1 customer arrives in a minute. This is the same as finding the probability that 0 or 1 customer arrives, and subtracting this from 1.

First, let's find P(X=0) and P(X=1).

P(X=0) = λ^0 * e^-λ / 0! = 0.2^0 * e^-0.2 / 1 = e^-0.2 = 0.81873

P(X=1) = λ^1 * e^-λ / 1! = 0.2^1 * e^-0.2 / 1 = 0.2 * e^-0.2 = 0.16375

The probability that 0 or 1 customer arrives is P(X=0) + P(X=1) = 0.81873 + 0.16375 = 0.98248.

Therefore, the probability that more than 1 customer arrives in a minute is 1 - P(X=0 or 1) = 1 - 0.98248 = 0.01752.

So, the probability that the shop has more than 1 arrival in a given minute is 0.018, when rounded to three decimal places.

Question

To solve this problem, we first need to find the average rate of arrivals per minute. This is given by the total number of arrivals divided by the total time. In this case, we have 17 customers over 85 minutes, so the average rate λ is 17/85 = 0.2 customers per minute.

The Poisson distribution gives the probability of a given number of events happening in a fixed interval of time, if these events happen with a known average rate and independently of the time since the last event. The formula for the Poisson distribution is:

P(X=k) = λ^k * e^-λ / k!

where:
- P(X=k) is the probability of k events,
- λ is the average rate of events per interval,
- e is the base of the natural logarithm (approximately 2.71828),
- k! is the factorial of k.

We want to find the probability that more than 1 customer arrives in a minute. This is the same as finding the probability that 0 or 1 customer arrives, and subtracting this from 1.

First, let's find P(X=0) and P(X=1).

P(X=0) = λ^0 * e^-λ / 0! = 0.2^0 * e^-0.2 / 1 = e^-0.2 = 0.81873

P(X=1) = λ^1 * e^-λ / 1! = 0.2^1 * e^-0.2 / 1 = 0.2 * e^-0.2 = 0.16375

The probability that 0 or 1 customer arrives is P(X=0) + P(X=1) = 0.81873 + 0.16375 = 0.98248.

Therefore, the probability that more than 1 customer arrives in a minute is 1 - P(X=0 or 1) = 1 - 0.98248 = 0.01752.

So, the probability that the shop has more than 1 arrival in a given minute is 0.018, when rounded to three decimal places.

Knowee AI · Accepted Answer