91Ó°ÊÓ

An environmental engineer measures the amount (by weight) of particulate pollution in air samples of a certain volume collected over two smokestacks at a coal-operated power plant. One of the stacks is equipped with a cleaning device. Let \(Y_{1}\) denote the amount of pollutant per sample collected above the stack that has no cleaning device and let \(Y_{2}\) denote the amount of pollutant per sample collected above the stack that is equipped with the cleaning device. Suppose that the relative frequency behavior of \(Y_{1}\) and \(Y_{2}\) can be modeled by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} k, & 0 \leq y_{1} \leq 2,0 \leq y_{2} \leq 1,2 y_{2} \leq y_{1} \\ 0, & \text { elsewhere } \end{array}\right.$$ That is. \(Y_{1}\) and \(Y_{2}\) are uniformly distributed over the region inside the triangle bounded by \(y_{1}=2,\) \(y_{2}=0,\) and \(2 y_{2}=y_{1}\) a. Find the value of \(k\) that makes this function a probability density function. b. Find \(P\left(Y_{1} \geq 3 Y_{2}\right)\). That is, find the probability that the cleaning device reduces the amount of pollutant by one-third or more.

Step by step solution

01

Understand the Problem

To solve this problem, we need to find the constant \(k\) that ensures the joint probability density function \(f(y_1, y_2)\) integrates to 1 over its domain. Then, find the probability \(P(Y_1 \geq 3Y_2)\).

02

Define the Domain

The domain is a triangular region bounded by the lines \(y_1 = 2\), \(y_2 = 0\), and \(2y_2 = y_1\). It can be described as \(0 \leq y_2 \leq 1\) and \(2y_2 \leq y_1 \leq 2\). This forms the triangular region over which we will integrate.

03

Set Up the Double Integral to Find \(k\)

The value of \(k\) is determined by ensuring the probability over the region sums to 1:\[\int_{0}^{1} \int_{2y_2}^{2} k \, dy_1 \, dy_2 = 1\]

04

Solve the Double Integral

Integrate with respect to \(y_1\):\[\int_{2y_2}^{2} k \, dy_1 = k(2 - 2y_2)\]Next, integrate with respect to \(y_2\):\[\int_{0}^{1} k(2 - 2y_2) \, dy_2 = k \left[ 2y_2 - y_2^2 \right]_0^1 \]\[= k[2(1) - 1^2 - (0)] = k[2 - 1] = k\]Set the integral equal to 1:\[k = 1\]Thus, \( k = 1 \), satisfying the probability condition.

05

Find \(P(Y_1 \geq 3Y_2)\)

We need to find the probability over the region where \(Y_1 \geq 3Y_2\) within the original triangular domain. The line \(y_1 = 3y_2\) intersects the domain at \(y_1 = 2y_2\) and \(y_1 = 2\) when \(y_2 = 2/3\).Integrate over this region:\[\int_{0}^{2/3} \int_{3y_2}^{2} 1 \, dy_1 \, dy_2\]First, integrate with respect to \(y_1\):\[\int_{3y_2}^{2} 1 \, dy_1 = (2 - 3y_2)\]Then integrate with respect to \(y_2\):\[\int_{0}^{2/3} (2 - 3y_2) \, dy_2 = \left[2y_2 - \frac{3}{2}y_2^2 \right]_0^{2/3}\]\[= (2\frac{2}{3} - \frac{3}{2}(\frac{2}{3})^2) = \frac{4}{3} - \frac{2}{3} = \frac{2}{3}\]Thus, \(P(Y_1 \geq 3Y_2) = \frac{2}{3}\).

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Key Concepts

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

Uniform Distribution

A uniform distribution is a type of probability distribution where every outcome in a given range is equally likely to occur. For the exercise given, the values of the pollutants above the smokestacks can take on are uniformly distributed over a specific triangular region.
This means that within this region, every pair of values \(Y_1\) and \(Y_2\) has the same probability of occurring.
The triangle is defined by the boundaries: \(y_1 = 2\), \(y_2 = 0\), and \(2y_2 = y_1\).
This sets the range where the uniform distribution applies.

• In our problem, each point within this triangle has an equal chance of being selected due to the constant probability density function value \(k\).
• Finding the value of \(k\) is essential in ensuring the function represents a proper probability distribution, which sums to 1 over its defined region.

Joint Probability

Joint probability in this context refers to the probability of two events occurring simultaneously: measuring a certain amount of pollution over time from two smokestacks at once.
The joint probability density function (pdf) for variables \(Y_1\) and \(Y_2\) describes how likely different combinations of these pollutant measurements are.
Since the distribution is uniform over a region, all combinations inside are equally likely, and outside this region, the probability is zero.

• To find total probabilities, one sums (or integrates in continuous cases) over the defined domain of possibilities.
• In our exercise, the function \(f(y_1, y_2)\) describes a joint pdf because it denotes probabilities for both \(Y_1\) and \(Y_2\) together.

Double Integral

A double integral is a mathematical tool used to compute the volume under a surface within a specific region.
In probability, it represents the summation of a probability density function over a two-dimensional area.For our problem, the double integral involved integrating \(f(y_1, y_2)\) over the specified triangular region to ensure the area (i.e., total probability) equals 1.
- The double integral was computed in two parts: first over \(y_1\), with boundaries from \(2y_2\) to \(2\), then over \(y_2\), from \(0\) to \(1\).- Solving the integral revealed that \(k=1\), thus normalizing the distribution.
This process guarantees that all probabilities within the domain add up according to the definition of a proper probability distribution.

Pollution Measurement

Pollution measurement in the context of this exercise focuses on determining how much particulate matter is present in air samples collected from smokestacks.
It involves quantifying pollutants with and without the effects of a cleaning device installed on one of the stacks, denoted as \(Y_1\) and \(Y_2\), respectively.
- The analysis looks at the cleaning device’s effectiveness by examining the probability \(P(Y_1 \geq 3Y_2)\). This represents the likelihood that pollution from the uncleaned stack is three times or more that from the cleaned one.

• Finding \(P(Y_1 \geq 3Y_2)\) requires examining the region of the joint distribution where this inequality holds.
• Through integral calculus over the defined regions, the answer \(\frac{2}{3}\) was obtained, indicating a significant chance that the cleaning device reduces pollutants by more than a third.