Question

Coffee and doughnuts. At a certain coffee shop, all the customers buy a cup of coffee; some also buy a doughnut. The shop owner believes that the number of cups he sells each day is normally distributed with a mean of 320 cups and a standard deviation of 20 cups. He also believes that the number of doughnuts he sells each day is independent of the coffee sales and is normally distributed with a mean of 150 doughnuts and a standard deviation of 12 a) The shop is open every day but Sunday. Assuming day-to-day sales are independent, what's the probability he'll sell over 2000 cups of coffee in a week? b) If he makes a profit of 50 cents on each cup of coffee and 40 cents on each doughnut, can he reasonably expect to have a day's profit of over \(\$ 300 ?\) Explain. c) What's the probability that on any given day he'll sell a doughnut to more than half of his coffee customers?

Short answer

a) \(\approx 5.16\%\), b) No, unlikely, c) \(\approx 23\%\)

Step-by-step solution

Define the Random Variable for Coffee Sales

The shop sells 320 cups of coffee daily on average, with a standard deviation of 20 cups. Thus, the daily coffee sales, \(C\), is modeled as \(C \sim N(320, 20^2)\).

Total Weekly Coffee Sales

The shop is open six days a week. Let \(W\) be the total weekly coffee sales. Since daily sales are independent, \(W = C_1 + C_2 + \ldots + C_6\), where each \(C_i \sim N(320,20^2)\). Therefore, \(W \sim N(1920, 2400)\).

Calculate Probability of Selling Over 2000 Cups

We need \(P(W > 2000)\). First, find the standard score: \(Z = \frac{2000 - 1920}{\sqrt{2400}} \approx 1.63\). Using the standard normal distribution, \(P(Z > 1.63) \approx 0.0516\).

Define Profit per Day

Profit per day comes from both coffee and doughnut sales: \(X = 0.5C + 0.4D\), where \(C \sim N(320, 20^2)\) and \(D \sim N(150, 12^2)\). Thus, \(X \sim N(188, 20^2 \times 0.5^2 + 12^2 \times 0.4^2) = N(188, 12.8^2)\).

Calculate Probability of Daily Profit Over $300

Find \(P(X > 300)\): \(Z = \frac{300 - 188}{12.8} \approx 8.75\). This \(Z\)-score is extremely high, indicating that \(P(X > 300) \approx 0\). Thus, over a reasonable period, exceeding a daily profit of \$300 is unlikely.

Define Doughnut Sales Relative to Coffee Sales

To sell a doughnut to more than half of the coffee customers: \(D > 0.5C\). If \(D - 0.5C > 0\), we define: \(Y = D - 0.5C\). \(Y \sim N(150 - 0.5 \times 320, \sqrt{12^2 + (0.5 \times 20)^2}) = N(-10, \sqrt{184})\).

Calculate Probability of High Doughnut Sales Ratio

Find \(P(Y > 0)\): \(Z = \frac{0 + 10}{\sqrt{184}} \approx 0.735\). Thus, \(P(Y > 0) \approx 0.23\). Therefore, there's a roughly 23% chance on any given day that more than half of the coffee customers will also buy a doughnut.

Key concepts

Normal Distribution

The concept of normal distribution is vital to many fields, including statistics, business, and social sciences. It describes how data values are distributed in a bell-shaped curve. This curve is symmetric, with most data points clustering around the mean, and the frequency of data points decreasing as you move away from the mean. In the context of the coffee shop exercise, both the daily sales of coffee and doughnuts are modeled as normal distributions, allowing for probabilistic predictions. Normal distributions are characterized by two parameters: the mean, which determines the center of the distribution, and the standard deviation, which describes the spread or variability of the data. The more data clusters near the mean, the smaller the standard deviation will be. This model is powerful for predicting sales, as real-world random variations often approximate this distribution.

Standard Deviation

Standard deviation is a measure of how spread out the numbers in a data set are. It's crucial for understanding the consistency, or variability, within a data set. A small standard deviation means that the numbers are close to the mean and to each other. Conversely, a large standard deviation means a wider spread of numbers. In the coffee shop scenario, the standard deviation helps the owner understand the variability in daily coffee and doughnut sales.

• Coffee Sales: A standard deviation of 20 cups indicates that on most days, sales will not deviate drastically from the average of 320 cups.
• Doughnut Sales: The standard deviation here is 12, showing a moderate variability around the mean sales of 150 doughnuts.

Understanding standard deviation helps the shop owner prepare for fluctuations in sales and manage inventory accordingly.

Random Variables

Random variables are numerical outcomes from a random phenomenon. They can be discrete, like the number of heads in coin flips, or continuous, like the amount of rainfall in a day. In the coffee shop problem, random variables are used to model daily sales. Specifically, the number of cups of coffee sold each day, represented by the variable \(C\), and the number of doughnuts sold daily, represented by \(D\), are random variables following a normal distribution. Each random variable has its own mean and standard deviation, providing the parameters for their respective normal distributions.Random variables help in creating mathematical models to represent and predict real-world processes. By understanding these variables, the coffee shop owner can estimate sales patterns over time, enabling better decision-making for stock and pricing.

Z-score Calculation

The Z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It is measured in terms of standard deviations from the mean. A Z-score tells you how many standard deviations a data point is from the mean, allowing you to standardize different data points from a normal distribution.

• A Z-score of 0 indicates that the data point is exactly at the mean.
• A positive Z-score signifies a value above the mean, while a negative Z-score indicates a value below the mean.

In the coffee shop scenario, calculating the Z-score helps determine the probability of selling certain numbers of coffee cups or making a desired profit. For instance, the shop owner uses a Z-score to calculate the probability of selling over 2000 cups of coffee in a week, allowing them to see how extreme or typical such a week would be in the context of their usual sales patterns. This is accomplished by comparing the actual sales figure to the expected quantity, measured in terms of standard deviation units from the mean, facilitating a standardized probability assessment of sales fluctuations.