Poisson Probability Problem
Use the Poisson distribution to calculate the probability of a given number of events occurring in a fixed interval of time.
Problem Scenario
A small coffee shop receives an average of 4 customers per 10-minute interval during the morning rush. Assuming customer arrivals follow a Poisson distribution, find: (a) the probability of exactly 6 customers arriving in a 10-minute period, and (b) the probability of at most 2 customers arriving in a 10-minute period.
Given Data
Requirements
- Apply the Poisson probability formula for part (a)
- Calculate the cumulative probability for part (b) by summing P(X=0), P(X=1), and P(X=2)
- Interpret the results
Solution
Step 1:
The Poisson probability formula is: P(X = k) = (e^(-lambda) * lambda^k) / k!, where lambda = 4.
Step 2:
Part (a): P(X = 6) = (e^(-4) * 4^6) / 6! = (0.01832 * 4096) / 720 = 75.05 / 720 = 0.1042.
Step 3:
Part (b): P(X <= 2) = P(X=0) + P(X=1) + P(X=2). P(X=0) = e^(-4) * 4^0 / 0! = 0.01832 * 1 / 1 = 0.01832. P(X=1) = e^(-4) * 4^1 / 1! = 0.01832 * 4 / 1 = 0.07326. P(X=2) = e^(-4) * 4^2 / 2! = 0.01832 * 16 / 2 = 0.14653.
Step 4:
Sum for part (b): P(X <= 2) = 0.01832 + 0.07326 + 0.14653 = 0.2381.
Step 5:
Interpretation: (a) There is about a 10.42% chance of exactly 6 customers arriving in any 10-minute window. (b) There is about a 23.81% chance of at most 2 customers arriving, which could help the shop plan staffing during slower periods.
Final Answer
(a) P(X = 6) = 0.1042 or about 10.4%. (b) P(X <= 2) = 0.2381 or about 23.8%. Exactly 6 arrivals is moderately unlikely, and having only 0, 1, or 2 customers in a 10-minute window occurs roughly a quarter of the time.
Key Takeaways
- ✓The Poisson distribution is ideal for modeling the number of events in a fixed interval when events occur independently at a constant average rate.
- ✓The mean and variance of a Poisson distribution are both equal to lambda. Here, both the mean and variance of customer arrivals are 4.
- ✓For cumulative probabilities like P(X <= k), sum the individual Poisson probabilities from X = 0 to X = k, or use complement rules when it is easier.
Common Errors to Avoid
- ✗Confusing lambda with k. Lambda is the average rate (parameter of the distribution), while k is the specific number of events you are calculating the probability for.
- ✗Forgetting to compute the factorial correctly, especially for larger k values. Recall that 0! = 1, 1! = 1, 2! = 2, 3! = 6, etc.
- ✗Using the Poisson distribution when events are not independent or the rate is not constant over the interval. In those cases, consider other models.
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Common questions about this problem type
The Poisson distribution is a limiting case of the binomial distribution when n is very large, p is very small, and the product np = lambda is moderate. It is often used as an approximation for the binomial when n >= 100 and p <= 0.01.
Adjust lambda proportionally. If the rate is 4 per 10 minutes and you want a 30-minute interval, use lambda = 4 * 3 = 12. The Poisson assumption of a constant rate allows this scaling as long as the rate truly remains constant over the longer interval.