Problem 5 - 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) \(E(X)\). $$ \begin{array}{l|llll} x_{i} & 1 & 2 & 3 & 4 \\ \hline p_{i} & 0.4 & 0.2 & 0.2 & 0.2 \end{array} $$

Step-by-Step Solution

Answer

(a) 0.6; (b) 2.2

Step 1: Understanding the Probability Distribution

The table provides a discrete probability distribution for the random variable \(X\). Each \(x_i\) is a possible outcome, and each \(p_i\) is the probability of that outcome. Here, \(X\) can be 1, 2, 3, or 4 with respective probabilities 0.4, 0.2, 0.2, and 0.2.

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

To find \(P(X \geq 2)\), sum the probabilities of \(X\) being 2, 3, or 4. So, \(P(X \geq 2) = P(X=2) + P(X=3) + P(X=4) = 0.2 + 0.2 + 0.2 = 0.6\).

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

The expected value \(E(X)\) is calculated using the formula \(E(X) = \sum (x_i \cdot p_i)\). Compute each term as follows: \(1 \times 0.4 = 0.4\), \(2 \times 0.2 = 0.4\), \(3 \times 0.2 = 0.6\), \(4 \times 0.2 = 0.8\). Add these values to find \(E(X)\): \(0.4 + 0.4 + 0.6 + 0.8 = 2.2\).

Key Concepts

Understanding Random Variables

Random variables are quite fascinating in probability theory. They are basically numeric representations of outcomes of a random phenomenon. A random variable, like the *X* in our problem, can take on a set of possible values, each associated with a probability.
To clarify further, imagine tossing a four-sided die where each side is labeled with a number. Here, the outcome of a toss is your random variable. For each possible rolled number, you get a different value.
In problems involving probability distributions, the random variable helps us focus on the numeric outcomes, allowing for mathematical analysis and calculation of probabilities.

Decoding Expected Value

The expected value is a key concept in determining the average outcome of a random variable, over many repetitions of an experiment. It's like finding the average score you might expect if you rolled that four-sided die many times.
The formula for expected value is: \ E(X) = \sum (x_i \cdot p_i) \, where each outcome value \(x_i\) is multiplied by its corresponding probability \(p_i\), and then summed together.
In simple terms:

• Multiply each possible outcome by its probability.
• Add all these products together.

Using our example, \(E(X) = 1 \times 0.4 + 2 \times 0.2 + 3 \times 0.2 + 4 \times 0.2 = 2.2\). This means that, on average, the outcome would stabilize around 2.2 over a large number of trials.

Probability Distribution Simplified

A probability distribution provides a comprehensive overview of all possible values of a random variable and their associated probabilities. It's like having a complete roadmap of outcomes directed by chance.
In discrete probability distributions, which deal with finite sets of outcomes, every possible value of the random variable is paired with a probability. These probabilities must sum to 1, ensuring that every conceivable outcome is accounted for.
Consider our simple example:

• Value 1 has a 0.4 chance.
• Values 2, 3, and 4 each have 0.2 chance.

The sum is: \(0.4 + 0.2 + 0.2 + 0.2 = 1\). This total of 1 ensures that one of those values must occur, providing a clear picture of the likelihood of each event happening.