Problem 7 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Based on past experience, the manager of the VideoRama Store has compiled the following table, which gives the probabilities that a customer who enters the VideoRama Store will buy \(0,1,2,3\), or 4 DVDs. How many DVDs can a customer entering this store be expected to buy? $$\begin{array}{lccccc} \hline \text { DVDs } & 0 & 1 & 2 & 3 & 4 \\\\\hline \text { Probability } & .42 & .36 & .14 & .05 & .03 \\ \hline\end{array}$$

Step-by-Step Solution

Answer

A customer entering the VideoRama Store can be expected to buy approximately \(0.91\) DVDs.

Step 1: Understand the concept of the expected value

The expected value is a way to find the average outcome of a random event when considering all possible outcomes and their probabilities. In this scenario, the random event is the number of DVDs a customer will buy. We are given probabilities for each outcome (0, 1, 2, 3, or 4 DVDs). We will multiply each outcome by its probability, and sum up the results.

Step 2: Calculate the expected number of DVDs bought

To find the expected number of DVDs a customer will buy, we'll now multiply each outcome (number of DVDs) by its respective probability and sum up all the results. Please see the calculations below: Expected number of DVDs = (0 × P(0)) + (1 × P(1)) + (2 × P(2)) + (3 × P(3)) + (4 × P(4)) where P(x) denotes the probability of buying x DVDs. Expected number of DVDs = (0 × 0.42) + (1 × 0.36) + (2 × 0.14) + (3 × 0.05) + (4 × 0.03)

Step 3: Evaluate the expression

We'll now plug the probabilities into the expected number of DVDs formula: Expected number of DVDs = (0 × 0.42) + (1 × 0.36) + (2 × 0.14) + (3 × 0.05) + (4 × 0.03) = 0 + 0.36 + 0.28 + 0.15 + 0.12 Now, we sum up all the values: Expected number of DVDs = 0.36 + 0.28 + 0.15 + 0.12 = 0.91

Step 4: Interpret the result

The expected number of DVDs a customer will buy when entering the store is 0.91. It means that, on average, each customer entering the store will purchase 0.91 DVDs - which is slightly less than one DVD per customer.

Key Concepts

Probability Distribution

A probability distribution provides a comprehensive framework that outlines all possible outcomes of a random event and their associated probabilities. In the context of the VideoRama Store exercise, the probability distribution reflects the likelihood of a customer purchasing between 0 to 4 DVDs.

This distribution is crucial because it allows us to see all possible scenarios and understand their relative chances. Here, each outcome—say, buying 0 DVDs versus 2 DVDs—has a specific probability tied to it, like 0.42 for 0 DVDs or 0.14 for 2 DVDs.

Probability distributions come in different types, such as discrete and continuous, based on the nature of the outcomes. In our case, we have a discrete probability distribution since the number of DVDs is countable and distinct (0, 1, 2, 3, or 4).

Key points to remember about probability distributions:

• The sum of all probabilities in a distribution must equal 1. This confirms that one of the described outcomes will occur.
• Each individual probability measures the chance of a specific outcome appearing in a single trial or observation.

Understanding this concept sets the stage for calculating the expected value, which effectively summarizes the distribution into a cohesive figure.

Random Variable

A random variable is a fundamental concept in probability and statistics that assigns numerical values to outcomes of a random process. In simpler terms, it's a variable whose possible values result from a random phenomenon.

In the VideoRama scenario, the random variable, often denoted as \( X \), represents the number of DVDs a customer purchases during a visit. This variable can take on one of the integer values: 0, 1, 2, 3, or 4.

Key features of random variables include:

• Discrete Random Variables: These have a countable number of possible values. Our scenario is an example where customers can buy a specific number of DVDs, making it discrete.
• Probability Distribution Association: Each random variable has a probability distribution that defines the likelihood of its various outcomes.

The purpose of a random variable is to bridge the gap between the real-world context of a problem and the mathematical calculations needed to draw insights, like calculating the expected value.

Mathematical Expectation

Mathematical expectation, also known as expected value, is a core concept that describes the anticipated average outcome over many repetitions of a random process. It provides a single summary measure for the random variable by incorporating both the possible values it can take and their respective probabilities.

In the VideoRama exercise, the mathematical expectation calculated is the expected number of DVDs a customer might buy. This is done by weighing each potential number of DVDs by how likely it is to happen and then summing these weighted outcomes. The actual calculation looks like this:

\[ E(X) = 0 \times 0.42 + 1 \times 0.36 + 2 \times 0.14 + 3 \times 0.05 + 4 \times 0.03 = 0.91 \]

This result—0.91 DVDs—indicates that, on average, a customer purchases less than one DVD per visit, pointing to the customers' overall purchasing behavior.

• The expected value gives us an average, but this doesn't mean any single customer will buy 0.91 DVDs.
• It is invaluable in making data-driven decisions and understanding trends over time.

Understanding mathematical expectation will enable you to predict outcomes more skillfully and apply these insights in practical scenarios.