- Cafe Espresso is a very popular coffee shop at UCI. The owner of Cafe Espresso notices that on average 85 customers order coffee on a given day. Let X be the number of customers who order coffee at Cafe Espresso, and we will assume that
. (a) Assume that the expected value of X is a constant daily rate and the number of customers who order coffee at Cafe Espresso is independent of any other day. What is the distribution of X? Justify your answer. (b) Using the distribution from part (a), what is the Mean and Standard Deviation of the number of customers who order coffee at cafe express in a give day. (c) Using the distributing from part (a), what is the probability that at least 81 customers order coffee at Cafe Espresso in a given day? (d) Suppose that the number of costumers who order coffee at Cafe Espresso in any increment of time is independent of the number of costumer who order coffee at Cafe Espresso in any other increment of time. Suppose that Cafe Espresso only opens between 6:00am and 3:00pm. 1) What is the probability that at least 11 costumer order coffee in one hour during a given day? 2) What is the probability that no costumer order coffee in one hour during a given day? (e) Using the assumption from above, what is the expected number of customers who order coffee at Cafe Espresso in one quarter? Assume that each day is independent of the other days and that there are 70 days in a quarter.
The Correct Answer and Explanation is:
(a) What is the distribution of X?
The number of customers XXX who order coffee per day is a count of events (customer arrivals) that:
- Happen independently from day to day.
- Happen at a constant average rate (85 per day).
- Cannot be negative and take only whole number values.
This is characteristic of a Poisson distribution.
So,
Answer: X∼Poisson(λ=85)X \sim \text{Poisson}(\lambda = 85)X∼Poisson(λ=85),
where λ=85\lambda = 85λ=85 is the average number of customers per day.
(b) Mean and Standard Deviation of X
For a Poisson distribution:
- The mean is μ=λ=85\mu = \lambda = 85μ=λ=85.
- The standard deviation is σ=λ=85≈9.22\sigma = \sqrt{\lambda} = \sqrt{85} \approx 9.22σ=λ=85≈9.22.
(c) Probability that at least 81 customers order coffee in a day
We use the Poisson distribution: P(X≥81)=1−P(X≤80)P(X \geq 81) = 1 – P(X \leq 80)P(X≥81)=1−P(X≤80)
Because λ=85\lambda = 85λ=85 is large, we approximate the Poisson distribution with a normal distribution: X∼N(85,9.222)X \sim N(85, 9.22^2)X∼N(85,9.222)
Using a continuity correction: P(X≥81)≈P(Z≥80.5−859.22)=P(Z≥−0.49)P(X \geq 81) \approx P\left(Z \geq \frac{80.5 – 85}{9.22}\right) = P(Z \geq -0.49)P(X≥81)≈P(Z≥9.2280.5−85)=P(Z≥−0.49)
Using standard normal table: P(Z≥−0.49)≈0.6879P(Z \geq -0.49) \approx 0.6879P(Z≥−0.49)≈0.6879
So,
Answer: P(X≥81)≈0.688\boxed{P(X \geq 81) \approx 0.688}P(X≥81)≈0.688
(d) Cafe Espresso is open 9 hours per day (6:00am–3:00pm).
Average per hour: λhour=859≈9.44\lambda_{\text{hour}} = \frac{85}{9} \approx 9.44λhour=985≈9.44
1) Probability at least 11 customers in one hour:
Use Poisson approximation: P(X≥11)=1−P(X≤10)P(X \geq 11) = 1 – P(X \leq 10)P(X≥11)=1−P(X≤10)
Using a Poisson calculator or table with λ=9.44\lambda = 9.44λ=9.44: P(X≤10)≈0.7212⇒P(X≥11)≈1−0.7212=0.2788P(X \leq 10) \approx 0.7212 \Rightarrow P(X \geq 11) \approx 1 – 0.7212 = 0.2788P(X≤10)≈0.7212⇒P(X≥11)≈1−0.7212=0.2788
Answer: P(X≥11)≈0.279\boxed{P(X \geq 11) \approx 0.279}P(X≥11)≈0.279
2) Probability no customer in one hour:
P(X=0)=e−9.44⋅9.4400!=e−9.44≈8.0×10−5P(X = 0) = e^{-9.44} \cdot \frac{9.44^0}{0!} = e^{-9.44} \approx 8.0 \times 10^{-5}P(X=0)=e−9.44⋅0!9.440=e−9.44≈8.0×10−5
Answer: P(X=0)≈0.00008\boxed{P(X = 0) \approx 0.00008}P(X=0)≈0.00008
(e) Expected number of customers in a quarter (70 days)
If each day is independent and λ=85\lambda = 85λ=85, for 70 days: Expected total customers=70×85=5950\text{Expected total customers} = 70 \times 85 = \boxed{5950}Expected total customers=70×85=5950
Explanation
The number of customers per day at Cafe Espresso follows a Poisson distribution because it represents counts of independent events (customer visits) occurring at a constant average rate. The Poisson distribution is ideal for modeling the number of events in fixed intervals of time when those events occur independently and randomly. Given the average is 85 customers per day, we say X∼Poisson(85)X \sim \text{Poisson}(85)X∼Poisson(85).
The mean of a Poisson distribution is equal to its parameter λ\lambdaλ, so the expected number of customers is 85. The standard deviation is the square root of λ\lambdaλ, so 85≈9.22\sqrt{85} \approx 9.2285≈9.22.
To find the probability of having at least 81 customers in a day, we approximate the Poisson with a normal distribution (since 85 is large) and apply a continuity correction. This gives us a probability of about 68.8%.
Next, since the shop operates for 9 hours daily, we divide the average rate to get about 9.44 customers per hour. Using this hourly Poisson rate, we find the probability of having at least 11 customers in one hour is approximately 27.9%, and the probability of no customers is extremely low—about 0.008%.
Finally, over 70 independent days in a quarter, the expected total number of customers is simply 70×85=595070 \times 85 = 595070×85=5950. This illustrates how Poisson processes can be extended to longer time frames while maintaining independence and a constant rate.
