91Ó°ÊÓ

In Exercise \(9,\) the distribution came "out of a hat." In this problem, we will again consider an experiment whose outcomes are not equally likely. We will determine a function \(f(x)\) which can be used to determine the probability of certain events. Let \(T\) be the right triangle in the plane with vertices at the points \((0,0),(1,0),\) and \((0,1) .\) The experiment consists of picking a point at random in the interior of \(T\), and recording only the \(x\) -coordinate of the point. Thus, the sample space is the set [0,1] , but the outcomes do not seem to be equally likely. We can simulate this experiment by asking a computer to return two random real numbers in \([0,1],\) and recording the first of these two numbers if their sum is less than \(1 .\) Write this program and run it for 10,000 trials. Then make a bar graph of the result, breaking the interval [0,1] into 10 intervals. Compare the bar graph with the function \(f(x)=2-2 x\). Now show that there is a constant \(c\) such that the height of \(T\) at the \(x\) -coordinate value \(x\) is \(c\) times \(f(x)\) for every \(x\) in \([0,1] .\) Finally, show that $$ \int_{0}^{1} f(x) d x=1 $$ How might one use the function \(f(x)\) to determine the probability that the outcome is between .2 and \(.5 ?\)

Step by step solution

01

Understanding the Experiment

The experiment involves choosing a point at random within a right triangle defined by the vertices (0,0), (1,0), and (0,1). We only record the x-coordinate of this point, resulting in a sample space ranging from 0 to 1. However, each possible x-value doesn't have an equal likelihood due to the triangular shape.

02

Simulating the Experiment

To simulate the experiment, we generate two random numbers, y1 and y2, between 0 and 1. We then record y1 if the sum y1+y2 is less than 1. This efficiently simulates picking a point within the triangle and obtaining its x-coordinate.

03

Defining the Probability Density Function

The function f(x)=2-2x represents a probability density function for the x-coordinate because it decreases linearly from 2 down to 0 as x increases from 0 to 1. It models the gradually diminishing likelihood of an x-value as we approach the base of the triangle from the origin.

04

Calculate Proportionality Constant

The height of the triangle at any point x can be written as 1-x since the line connecting (1,0) and (0,1) is y=1-x. We need a constant c such that 1-x = c*(2-2x) to adjust for the difference in likelihood. Solving yields c=1/2.

05

Verify the Integral of f(x)

To ensure f(x)=2-2x is a valid probability density function, we need the integral over its defined range [0,1] to equal 1. Calculate the integral: \[\int_{0}^{1} (2-2x) \, dx = [2x - x^2]_0^1 = (2\times1 - 1^2) - (0) = 1\] which confirms the function is valid.

06

Using f(x) for Probability Calculations

The function f(x) can be used to find probabilities over specific intervals by integrating over that interval. To find the probability that the outcome is between 0.2 and 0.5, compute:\[\int_{0.2}^{0.5} (2-2x) \, dx = [2x - x^2]_{0.2}^{0.5} = (2\times0.5 - 0.5^2) - (2\times0.2 - 0.2^2) = 0.81\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Triangular Distribution

The Triangular Distribution is a fascinating concept in probability theory, providing a simple model for variables that are bound between a lower and upper limit, where a peak occurs at a particular value. It is defined by three parameters: the minimum value, the maximum value, and the peak (mode) where the highest probability density occurs.

In our specific exercise, the triangular distribution is applied to the random selection of points from a right triangle with vertices (0,0), (1,0), and (0,1). This triangle represents an experiment where we examine the probability distribution solely based on the x-coordinate.

The distribution is not uniform because the likelihood changes as you move along the base of the triangle, making some areas more probable than others. Such a triangular distribution is often visualized with the highest probability density at one end (in this case, the point at (0,0)) and tapering off at the opposite end.

Random Variables

Random variables are essential elements in probability, capturing the outcomes of a probabilistic experiment numerically. They can be thought of as functions that assign numbers to outcomes of a random process.

In the context of our exercise, the random variable of interest is the x-coordinate of a randomly chosen point from the triangle. Instead of considering points as two-dimensional coordinates (x, y), our experiment simplifies matters by reducing focus to just the x-coordinate.

This random variable does not distribute evenly, given the triangular nature, so each possible x-value within the range \([0,1]\) is not equally likely. This intrinsic characteristic influences how probabilities are calculated within the context and needs careful consideration when analyzing or simulating the experiment.

Probability Calculations

Probability calculations allow us to determine the likelihood of various outcomes within a defined range, particularly by leveraging the probability density function (PDF).

For a continuous random variable like in our exercise, probabilities are found by integrating the PDF over the desired interval. The function given in this context is \(f(x) = 2 - 2x\), which decreases linearly and identifies the density for each x-value across [0,1].

• Defining the PDF: The function \(f(x)\) serves as the PDF, illustrating how probability densities vary along the x-axis.
• Calculating Specific Intervals: To find the probability of the x-coordinate occurring within a specific interval, such as between 0.2 and 0.5, integrate \(f(x)\) over this range to yield the probability.

This method ensures that even non-uniform distributions like our triangular one can still be effectively analyzed.

Integration

Integration is a fundamental mathematical tool used for finding areas under curves, which directly relates to calculating probabilities in continuous distributions.

For our triangular distribution, integrating the PDF \(f(x) = 2 - 2x\) across the interval [0,1] verifies it as a valid PDF since the integral equals 1. This result confirms that the entire probability space is accurately accounted for.

• Basic Integration: Calculate \(\int_{0}^{1} (2-2x) \, dx\) using fundamental integration techniques, ensuring that the outcome equals 1.
• Interval Specific Integration: For probabilities between specific values, such as 0.2 to 0.5, perform the definite integral over the relevant bounds to derive the specific probability.

Integration within this context solidifies our understanding of how probability densities function across continuous random variables, playing a vital role in precise probability assessment.