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) 0.1; (b) 0.

Step 1: Define the Probability for X ≥ 2

To find \( P(X \geq 2) \), we need to sum the probabilities for all \( x \) values that are greater than or equal to 2. By looking at the table, we find that only \( x = 2 \) satisfies this condition with a probability of 0.1.

Step 2: Calculate P(X ≥ 2)

Since \( P(X \geq 2) = P(X = 2) \), we simply use the value from the table:\\[ P(X \geq 2) = p(2) = 0.1 \]

Step 3: Define Expected Value Formula

The expected value \( E(X) \) is calculated using the formula: \\[ E(X) = \sum_{i} x_{i} p_{i} \] \where \( x_{i} \) are the possible values of the random variable and \( p_{i} \) are their corresponding probabilities.

Step 4: Calculate Each Term for E(X)

Compute the product of each \( x_{i} \) and \( p_{i} \): \\( -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 \)

Step 5: Sum the Terms for E(X)

Add the results from Step 4 to find \( E(X) \): \\[ E(X) = -0.2 + (-0.2) + 0 + 0.2 + 0.2 = 0 \]

Key Concepts

Random Variables

When talking about probability, random variables are a key concept. A random variable is essentially a function that assigns a numerical value to each possible outcome of a random phenomenon. Think of it as a way to quantify uncertainty.

• In essence, random variables can be discrete or continuous. **Discrete random variables** have specific values, like roll results on a die (1 through 6).
• In our exercise, the random variable \( X \) takes on discrete values of \(-2, -1, 0, 1, \) and \( 2 \).

The random variable is accompanied by what is known as a discrete probability distribution. This distribution lists all possible outcomes and their associated probabilities. It helps in understanding how likely each outcome is, providing a systematic approach to dealing with randomness.

Expected Value

The expected value is like the average of all possible outcomes, weighted by their probabilities. It's a crucial concept in probability theory as it provides a measure of the center of the distribution.

• To find the expected value \( E(X) \), we use the formula \( E(X) = \sum_{i} x_{i} p_{i} \), where \( x_{i} \) represents the values the random variable can take, and \( p_{i} \) their probabilities.
• In the exercise, this is done by multiplying each outcome \( x_{i} \) by its probability \( p_{i} \) and summing these products: \( -0.2, -0.2, 0, 0.2, \) and \( 0.2 \).

After calculating, you get a result that represents a "weighted average" of all outcomes, which in this case is zero. It might seem surprising, but it shows that, on average, this distribution achieves a balance between negative and positive values.

Probability Theory

Probability theory is the branch of mathematics concerned with analyzing random phenomena. It provides a mathematical foundation to talk about uncertainty and outcomes. One of the foundational tools it provides is the probability distribution.

• A probability distribution must satisfy two main conditions: probabilities are between 0 and 1, and the sum of probabilities equals 1.
• Our given distribution with \( p_{i} = 0.1, 0.2, 0.4, 0.2, \) and \( 0.1 \) successfully sums to 1, confirming it's a valid distribution.

This theory also helps in calculating specific probabilities, such as finding \( P(X \geq 2) \). From the exercise, we calculate this by identifying all outcomes that meet the condition \( X \geq 2 \) and summing their probabilities. Here, it's as simple as recognizing the single outcome when \( x = 2 \), having a probability of 0.1. Such calculations highlight probability theory's utility in transforming randomness into informed estimates.