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 E[X] = 85.
(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 :
Let’s address each part of the question step by step:
(a) What is the distribution of X? Justify your answer.
Given that E[X] = 85, and the number of customers ordering coffee at Cafe Espresso is assumed to be independent of any other day, we can assume that X follows a Poisson distribution. The Poisson distribution is often used to model the number of occurrences of an event in a fixed interval of time or space, provided these events occur independently and at a constant average rate.
In this case, the average number of customers ordering coffee per day is 85. The Poisson distribution is characterized by the parameter λ, which is the rate of occurrence (in this case, the expected number of customers per day). Therefore, the number of customers ordering coffee at Cafe Espresso follows a Poisson distribution with λ = 85.
Thus, X ~ Poisson(85).
(b) Mean and Standard Deviation of X.
For a Poisson distribution, the mean and variance are both equal to λ. Therefore:
- Mean of X = λ = 85
- The standard deviation is the square root of the variance, which is also λ. Hence:
$$
\text{Standard Deviation} = \sqrt{\lambda} = \sqrt{85} \approx 9.22
$$
So:
- Mean = 85
- Standard Deviation ≈ 9.22
(c) Probability that at least 81 customers order coffee.
We need to find $P(X \geq 81)$ where X ~ Poisson(85).
To calculate this, we use the cumulative distribution function (CDF) for the Poisson distribution. Specifically, we find:
$$
P(X \geq 81) = 1 – P(X < 81) = 1 – P(X \leq 80)
$$
This can be computed using a Poisson CDF calculator or statistical tables. Using a calculator for Poisson(85), we find:
$$
P(X \geq 81) \approx 1 – P(X \leq 80) \approx 0.7408
$$
So, the probability that at least 81 customers order coffee is approximately 0.7408, or 74.08%.
(d) 1) Probability that at least 11 customers order coffee in one hour.
Here, we assume that the number of customers who order coffee follows a Poisson distribution over a smaller time frame (one hour). Since the total expected number of customers per day is 85, we can compute the expected number of customers per hour as:
$$
\lambda_{\text{hour}} = \frac{85}{9} \approx 9.44
$$
Thus, the number of customers ordering coffee in one hour follows a Poisson distribution with λ = 9.44.
Now, we want to find the probability that at least 11 customers order coffee in one hour:
$$
P(X \geq 11) = 1 – P(X \leq 10)
$$
Using a Poisson CDF calculator for Poisson(9.44):
$$
P(X \geq 11) \approx 1 – P(X \leq 10) \approx 0.4502
$$
So, the probability that at least 11 customers order coffee in one hour is approximately 0.4502 or 45.02%.
(d) 2) Probability that no customer orders coffee in one hour.
To find the probability that no customer orders coffee in one hour, we need to calculate $P(X = 0)$ for X ~ Poisson(9.44):
The probability mass function (PMF) for a Poisson distribution is given by:
$$
P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}
$$
For k = 0, this simplifies to:
$$
P(X = 0) = e^{-\lambda} = e^{-9.44} \approx 7.83 \times 10^{-5}
$$
So, the probability that no customer orders coffee in one hour is approximately 0.0000783, or 0.00783%.
(e) Expected number of customers in one quarter.
To find the expected number of customers who order coffee in one quarter, we first calculate the expected number of customers per day, which is 85. Since the quarter consists of 70 days, the total expected number of customers in the quarter is:
$$
E[\text{Total Customers in Quarter}] = 85 \times 70 = 5950
$$
Thus, the expected number of customers who order coffee at Cafe Espresso in one quarter is 5950.
Conclusion:
- (a) The distribution of X is Poisson(85).
- (b) The mean of X is 85, and the standard deviation is approximately 9.22.
- (c) The probability that at least 81 customers order coffee is approximately 74.08%.
- (d) The probability that at least 11 customers order coffee in one hour is approximately 45.02%, and the probability that no customer orders coffee in one hour is approximately 0.00783%.
- (e) The expected number of customers in one quarter is 5950.