Problem 4 - 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 \((b) E(X)\). $$ \begin{array}{l|lllll} x_{i} & -2 & -1 & 0 & 1 & 2 \\ \hline p_{i} & 0.1 & 0.2 & 0.4 & 0.2 & 0.1 \end{array} $$

Step-by-Step Solution

Answer

(a) \( P(X \geq 2) = 0.1 \); (b) \( E(X) = 0.0 \).

Step 1: Identify Probabilities for P(X ≥ 2)

To find \( P(X \geq 2) \), we first identify the probability associated with the value \( x = 2 \). According to the table, \( P(X = 2) = 0.1 \). Since no other values are greater than 2, \( P(X \geq 2) = P(X = 2) = 0.1 \).

Step 2: Calculate Expected Value E(X) - Identify Components

The expected value of a discrete random variable \( X \) is calculated using the formula \( E(X) = \sum_{i} x_{i} p_{i} \). We need to find the product \( x_{i} \times p_{i} \) for each value in our table and sum these products.

Step 3: Compute Products for E(X)

Compute for each \( i \):- \( x_{1} = -2 \), \( p_{1} = 0.1 \), so \( x_{1}p_{1} = -2 \times 0.1 = -0.2 \).- \( x_{2} = -1 \), \( p_{2} = 0.2 \), so \( x_{2}p_{2} = -1 \times 0.2 = -0.2 \).- \( x_{3} = 0 \), \( p_{3} = 0.4 \), so \( x_{3}p_{3} = 0 \times 0.4 = 0 \).- \( x_{4} = 1 \), \( p_{4} = 0.2 \), so \( x_{4}p_{4} = 1 \times 0.2 = 0.2 \).- \( x_{5} = 2 \), \( p_{5} = 0.1 \), so \( x_{5}p_{5} = 2 \times 0.1 = 0.2 \).

Step 4: Sum Products to Find E(X)

Now sum the products from Step 3: \(-0.2 + (-0.2) + 0 + 0.2 + 0.2 = 0.0\). Thus, the expected value \( E(X) = 0.0 \).

Key Concepts

Random Variable

In probability theory, a random variable is a fundamental concept that allows us to map outcomes of a random process to numbers. A **discrete random variable** can take on a finite or countably infinite number of distinct values, each associated with a probability. For instance, the random variable \(X\) in our problem takes on the values \(-2, -1, 0, 1,\) and \(2\). Each value has a corresponding probability, which is given in the probability distribution table.

Understanding random variables is crucial because they connect the abstract notion of randomness to mathematically analyzable models. In practical terms, whenever you're dealing with quantities like the outcome of a dice roll, the number of defective items in a sample, or the number of heads in a series of coin tosses, you're dealing with a discrete random variable.

Remember, the values that a random variable can take are known as its support. Here, the support of \(X\) is \{-2, -1, 0, 1, 2\}.

Expected Value

The **expected value** of a random variable is like the center of its distribution, providing a sense of the variable's average outcome over many trials. For a discrete random variable, the expected value \(E(X)\) is computed as:\[E(X) = \sum_{i} x_{i} p_{i}\]This equation tells us to multiply each value \(x_{i}\) of the random variable \(X\) by its probability \(p_{i}\), then sum all those products.

In our exercise, we have calculated:

• \(-2 \times 0.1 = -0.2\)
• \(-1 \times 0.2 = -0.2\)
• \(0 \times 0.4 = 0\)
• \(1 \times 0.2 = 0.2\)
• \(2 \times 0.1 = 0.2\)

Summing these products gives us \(0.0\). This result means on average, the random variable \(X\) would produce an outcome of \(0.0\) if repeated multiple times.

The expected value reflects the probability-weighted average and doesn't always have to be a value from the variable's support.

Probability Theory

**Probability theory** is the branch of mathematics that deals with the analysis and interpretation of random phenomena. It's used to model the likelihood of different outcomes for events or processes that are inherently uncertain. In probability theory, the **discrete probability distribution** assigns probabilities to each possible value of a discrete random variable.

For our variable \(X\), we are provided with a probability distribution table that gives us the probabilities for each possible outcome. The sum of all these probabilities must equal \(1\). This is a crucial rule in all probability distributions: \[\sum p_{i} = 1\]where \(p_{i}\) are the probabilities associated with the outcomes.

Understanding and applying probability theory allows us to make informed predictions and decisions based on the likelihood of events. It forms the bedrock for many fields, from statistics and finance to machine learning and beyond. The exercise showcases classic applications of probability theory, helping you to calculate chances like \(P(X \geq 2)\) and determine expected outcomes. This foundational knowledge is vital for delving deeper into the world of data and uncertainty.