Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

A discrete probability distribution for a random variable \(X\) is given. Use the given distribution to find (a) \(P(X \geq 2)\) and (b) \(E(X)\). \begin{array}{l|llll} x_{i} & 0 & 1 & 2 & 3 \\ \hline p_{i} & 0.80 & 0.10 & 0.05 & 0.05 \end{array}

Step-by-Step Solution

Answer

(a) \(P(X \geq 2) = 0.10\); (b) \(E(X) = 0.35\).

Step 1: Understanding the Discrete Probability Distribution

A discrete probability distribution lists the probability associated with each possible outcome of a random variable. In this exercise, we have the random variable \(X\) which takes on the values 0, 1, 2, or 3 with corresponding probabilities 0.80, 0.10, 0.05, and 0.05, respectively.

Step 2: Calculating \(P(X \geq 2)\)

To find \(P(X \geq 2)\), we sum the probabilities of \(X\) being 2 and 3. Thus, \(P(X \geq 2) = P(X = 2) + P(X = 3) = 0.05 + 0.05 = 0.10\).

Step 3: Finding the Expected Value \(E(X)\)

The expected value \(E(X)\) is calculated by multiplying each outcome by its probability and summing these products: - For \(x = 0\), \(0 \times 0.80 = 0\)- For \(x = 1\), \(1 \times 0.10 = 0.10\)- For \(x = 2\), \(2 \times 0.05 = 0.10\)- For \(x = 3\), \(3 \times 0.05 = 0.15\)Summing these gives \(E(X) = 0 + 0.10 + 0.10 + 0.15 = 0.35\).

Key Concepts

Random Variable

A random variable, in simple terms, is a variable whose possible values are numerical outcomes of a random process. Each outcome is associated with a specific probability. In the context of the given exercise, we have a random variable \(X\) that can take on a set of defined values: 0, 1, 2, and 3.
The key aspect to understand is that these values represent potential occurrences in a probabilistic event, like rolling a die or picking a card. What makes a random variable distinct is how these values come into play in relation to the probabilities assigned to them.
These probabilities reflect the likelihood of each value occurring in a single instance of the random process. With the set values and associated probabilities, \(X\) offers us a way to quantify randomness in a structured way.

Expected Value

The expected value, often denoted as \(E(X)\), can be thought of as the average outcome you would expect if the random event represented by our random variable was repeated many times. When dealing with a discrete random variable such as \(X\), calculation of the expected value involves a weighted sum of all possible outcomes.

This means that each potential outcome value is multiplied by its corresponding probability, and then all those products are summed together.

• For \(x=0\), calculation is \(0 \times 0.80 = 0\).
• For \(x=1\), calculation is \(1 \times 0.10 = 0.10\).
• For \(x=2\), calculation is \(2 \times 0.05 = 0.10\).
• For \(x=3\), calculation is \(3 \times 0.05 = 0.15\).

Gathering all these values, we calculate \(E(X) = 0 + 0.10 + 0.10 + 0.15 = 0.35\).
This expected value tells us that, on average, we can expect an outcome of 0.35 when observing the random variable \(X\) across many trials.

Probability Calculation

Probability calculation in the realm of discrete probability distributions involves assessing the chance of the random variable attaining a certain value or range of values. In the provided scenario, we needed to find \(P(X \geq 2)\), which is the probability of the random variable \(X\) taking a value of 2 or more.
Calculation steps include:

• Identify the relevant events. Here, they are \(X = 2\) and \(X = 3\).
• Add the probabilities of these events. Specifically, \(P(X = 2) + P(X = 3) = 0.05 + 0.05 = 0.10\).

Thus, the probability that the random variable \(X\) equals or exceeds the value of 2 is 0.10. This calculation is essential for understanding how likely certain ranges of outcomes are within a given random process.

Probability Distribution Table

A probability distribution table is a handy tool that organizes a discrete set of outcomes associated with a random variable and displays them alongside their probabilities. It encapsulates the entire distribution of the variable in a concise format.
For instance, in the exercise given, the table presents values \(0, 1, 2,\) and \(3\) with their respective probabilities: 0.80, 0.10, 0.05, and 0.05.
This table is fundamental because:

• It offers a clear view of all possible outcomes and their associated chances, facilitating quick reference and easy interpretation.
• It serves as a starting point for further calculations or analysis, like determining expected values or specific probabilities.

A well-constructed probability distribution table is vital for anyone working with statistical data involving discrete random variables, as it lays the groundwork for thorough statistical analysis.