Problem 13 - Algebra 2

Question

Graph the probability distribution described by each function. $$ P(x)=\frac{2 x+1}{15} \text { for } x=1,2, \text { and } 3 $$

Step-by-Step Solution

Answer

The probability distribution can be graphed with points at (1,0.2), (2,0.33), and (3,0.47).

Step 1: Calculate P(1)

Substitute \( x = 1 \) into the function \( P(x) = \frac{2x + 1}{15} \). This gives: \( P(1) = \frac{2(1) + 1}{15} = \frac{3}{15} = 0.2 \)

Step 2: Calculate P(2)

Substitute \( x = 2 \) into the function \( P(x) = \frac{2x + 1}{15} \). This gives: \( P(2) = \frac{2(2) + 1}{15} = \frac{5}{15} = 0.33 \)

Step 3: Calculate P(3)

Substitute \( x = 3 \) into the function \( P(x) = \frac{2x + 1}{15} \). This gives: \( P(3) = \frac{2(3) + 1}{15} = \frac{7}{15} = 0.47 \)

Step 4: Graph the distribution

Draw a dot at (1,0.2), (2,0.33), and (3,0.47) corresponding to the calculated probabilities for each x value. As these are the only values x can take, the graph will consist of these three points

Key Concepts

Graphing Techniques

To create a visual representation of a probability distribution, you need to understand some basic graphing techniques. Here, the objective is to display the probabilities calculated for each value of the discrete random variable, which in this case, are 1, 2, and 3. Start by drawing a simple graph with a horizontal and vertical axis.

• The horizontal axis (x-axis) will represent the discrete values of the random variable. In our case, these values are 1, 2, and 3.
• The vertical axis (y-axis) corresponds to the probabilities associated with each value.

To plot the points:

• Firstly, locate the x-value on the x-axis. For example, begin with x = 1.
• Then, move vertically to the corresponding probability value on the y-axis. Here, P(1) = 0.2.
• Draw a dot at the intersection of these values. Repeat this process for x = 2 and x = 3, using their respective probabilities, 0.33 and 0.47.

These points will visually represent the probability distribution, clearly showing the likelihood of each outcome. This method allows students to easily see which outcomes are more or less likely compared to each other.

Discrete Probability

Discrete probability deals with events that have finite or countable outcomes. In this exercise, we are given a probability distribution for discrete random variable values of 1, 2, and 3. The function provided is:\[ P(x) = \frac{2x + 1}{15} \] The probability distribution function (PDF) gives us the likelihood of each possible outcome.
When dealing with discrete probabilities, the following properties are important:

• Each individual probability must be between 0 and 1.
• The total probability of all possible outcomes should add up to 1.

For this function:

• Calculate each probability: P(1), P(2), and P(3). These calculations yield 0.2, 0.33, and 0.47 respectively.
• If you add these probabilities together, they sum up to 1, confirming the function is indeed a valid probability distribution.

This provides a clear framework for determining how likely each individual outcome is and ensures that the function correctly represents a real-life scenario where probabilities are constrained by these rules.

Function Evaluation

Evaluation of the given function \( P(x) = \frac{2x + 1}{15} \) involves substituting specific values of \( x \), which represents the different outcomes we are interested in evaluating. This process gives us the probability for each possible outcome.
The steps to evaluate the function are straightforward:

• Substitute \( x = 1 \) into the function: \( P(1) = \frac{2(1) + 1}{15} = \frac{3}{15} = 0.2 \).
• Substitute \( x = 2 \) into the function: \( P(2) = \frac{2(2) + 1}{15} = \frac{5}{15} = 0.33 \).
• Substitute \( x = 3 \) into the function: \( P(3) = \frac{2(3) + 1}{15} = \frac{7}{15} = 0.47 \).

This method allows us to determine the likelihood of each discrete event occurring. It also demonstrates how a simple formula can be used to calculate probabilities for multiple outcomes efficiently. Function evaluation is a crucial aspect as it not only solves the problem but ensures the values align with theoretical expectations.