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How Many Cups of Coffee? Choose an adult age 18 or over in the Unaited States at random and ask, "How many cups of coffee do you drink on average per day?" Call the response \(X\) for short. Based on a large sample survey, here is a probability model for the answer you will get: \({ }^{7}\) \begin{tabular}{l|ccccc} \hline Number & 0 & 1 & 2 & 3 & 4 or more \\ \hline Probability & \(0.36\) & \(0.26\) & \(0.19\) & \(0.08\) & \(0.11\) \\ \hline \end{tabular} (a) Verify that this is a valid finite probability model. (b) Describe the event \(X<4\) in words. What is \(P(X<4)\) ? (c) Express the event "have at least one cup of coffee on an average day" in terms of \(X\). What is the probability of this event?

Step by step solution

01

Verify Validity of Probability Model

To verify the probability model, we need to check if the sum of the given probabilities equals 1. Calculate the sum: \[ 0.36 + 0.26 + 0.19 + 0.08 + 0.11 = 1.00 \] Since the total probability is 1, it is a valid model.

02

Describe Event X

Event \(X < 4\) means that a person drinks fewer than 4 cups of coffee per day. This includes the probabilities for drinking 0, 1, 2, or 3 cups of coffee.

03

Calculate P(X

To find \(P(X < 4)\), add the probabilities of drinking 0, 1, 2, or 3 cups: \[ P(X < 4) = 0.36 + 0.26 + 0.19 + 0.08 = 0.89 \] So, the probability of drinking fewer than 4 cups of coffee per day is 0.89.

04

Express and Calculate P(X≥1)

The event "have at least one cup of coffee" is written as \(X \geq 1\). To find \(P(X \geq 1)\), we subtract the probability of drinking 0 cups from 1: \[ P(X \geq 1) = 1 - P(X = 0) = 1 - 0.36 = 0.64 \] Therefore, the probability of having at least one cup of coffee is 0.64.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Calculation

Probability calculation is a fundamental concept in understanding how likely an event is to occur. This involves summing probabilities of single events to find the likelihood of a more complex event happening. In the coffee consumption example:

• We calculated the total probability of drinking fewer than 4 cups by adding probabilities for 0, 1, 2, and 3 cups.
• This resulted in a sum of 0.89, meaning there's an 89% chance that a randomly chosen adult drinks fewer than 4 cups.

Similarly, to determine the likelihood of drinking at least one cup, we subtracted the probability of not drinking any coffee (0 cups) from 1. This provided us with the probability of having at least one cup being 0.64 or 64%.

Making such calculations helps in understanding real-world behaviors and patterns, and allows effective decision-making based on these insights.

Finite Probability Model

A finite probability model involves a limited number of events and their associated probabilities. Each probability in such a model must lie between 0 and 1. Most importantly, the sum of all probabilities in the model should exactly equal 1. This ensures all possible outcomes have been considered and weighed appropriately.

• In our example, the number of cups ranges from 0 to 4 or more.
• Probabilities are given for each category, and when summed, give us 1.00, confirming the model's validity.

This type of model is often used in fields where outcomes are distinct and countable, ensuring comprehensive analysis by encompassing every potential scenario with a given probability.

Discrete Random Variables

In probability, a discrete random variable can take on only a specific list of values. These are well-defined and can be enumerated. For instance, the number of cups of coffee a person drinks daily is a discrete random variable.

• The random variable in our problem, denoted as \(X\), describes a person's coffee consumption.
• This variable can have values such as 0, 1, 2, 3, and 4 or more, each with an assigned probability.

Understanding discrete random variables allows us to use probability models to make predictions and assess the likelihood of different outcomes, facilitating more informed and data-driven explanations on behavioral studies and survey analyses.