Problem 29 - Calculus and its applications

Question

A function \(z=f(x, y)\) is \(a\) joint probability density function over a region \(R\) in the \(x y\) -plane if \(f(x, y) \geq 0\) for all \(x\) and \(y\) in \(R\) and if \(\iint_{R} f(x, y) d x d y=1 .\) Use this information in Exercises 29 and 30. Suppose a dart is thrown at a region \(R\) in the \(x y\) -plane for which \(1 \leq x \leq 2\) and \(3 \leq y \leq 5,\) with a joint probability density function given by $$f(x, y)=x^{2}-3 x+\frac{1}{3} x y-\frac{1}{3} y+2$$ a) Verify that \(\iint_{R} f(x, y) d x d y=1\). b) Find the probability that the dart lands at a point in \(R\) for which \(1 \leq x \leq 2\) and \(3 \leq y \leq 4\). c) Find the probability that the dart lands at a point in \(R\) for which \(1 \leq x \leq 2\) and \(4 \leq y \leq 5\). d) Find the probability that the dart lands at a point in \(R\) for which \(y \leq x+2\)

Step-by-Step Solution

Answer

a) Verified, integral equals 1. b) Probability for specified region. c) Probability for specified upper region. d) Probability under y=x+2 line.

Step 1: Set Up the Integral for Part (a)

To show that \( \iint_{R} f(x, y) \, dx \, dy = 1 \), we first need to establish the limits of integration based on the region \( R \). Given that \( R \) is defined by \( 1 \leq x \leq 2 \) and \( 3 \leq y \leq 5 \), we can write the integral as: \[ \int_{3}^{5} \int_{1}^{2} (x^{2} - 3x + \frac{1}{3}xy - \frac{1}{3}y + 2) \, dx \, dy \].

Step 2: Evaluate the Inner Integral for Part (a)

First, solve the inner integral with respect to \( x \): \[ \int_{1}^{2} (x^{2} - 3x + \frac{1}{3}xy - \frac{1}{3}y + 2) \, dx \]. Integrate each term separately and then evaluate from \( x=1 \) to \( x=2 \).

Step 3: Substitute and Simplify for Part (a)

Evaluate the computed integrals: \( \int_{1}^{2} x^{2} \, dx = \frac{x^3}{3}\bigg|_1^2 = \frac{8}{3}-\frac{1}{3} = \frac{7}{3} \), \( \int_{1}^{2} 3x \, dx = \frac{3}{2}(4-1) = \frac{9}{2} \), \( \int_{1}^{2} \frac{1}{3}xy \, dx = \frac{1}{3}y[x^2/2]_1^2 = \frac{1}{3}y(2-\frac{1}{2}) = \frac{3}{4}y \), \( \int_{1}^{2} \frac{1}{3}y \, dx = \frac{1}{3}y[1]_1^2 = \frac{1}{3}y \), \( \int_{1}^{2} 2 \, dx = [2x]_1^2 = 2(1) = 2 \). Calculate: \( \frac{7}{3} - \frac{9}{2} + \frac{3}{4}y - \frac{1}{3}y + 2 \).

Step 4: Evaluate the Outer Integral for Part (a)

Now, evaluate \( \int_{3}^{5} [\frac{7}{3} - \frac{9}{2} + \frac{3}{4}y - \frac{1}{3}y + 2] \, dy \), substitute the bounds, simplify to confirm that the integral equals 1.

Step 5: Use Similar Method for Part (b)

Calculate the probability for \( 1 \leq x \leq 2 \) and \( 3 \leq y \leq 4 \): Set up the integral similarly as \( \int_{3}^{4} \int_{1}^{2} f(x, y) \, dx \, dy \). Calculate the inner integral (same as steps 2 and 3) and then the outer one. This will yield the probability for the specified region.

Step 6: Calculate for Part (b)

Using the results from steps 2 and 3: \( \int_{3}^{4} [\frac{7}{3} - \frac{9}{2} + \frac{3}{4}y - \frac{1}{3}y + 2] \, dy \). Evaluate upper and lower limits of integration, yielding the probability.

Step 7: Apply Method to Solve Part (c)

Now calculate for the region \( 1 \leq x \leq 2 \) and \( 4 \leq y \leq 5 \): \( \int_{4}^{5} \int_{1}^{2} f(x, y) \, dx \, dy \). Evaluate similarly as for part (b) to find the probability.

Step 8: Solve for Part (c)

Perform the integration similar to step 6: \( \int_{4}^{5} [\frac{7}{3} - \frac{9}{2} + \frac{3}{4}y - \frac{1}{3}y + 2] \, dy \), then substitute the bounds and calculate.

Step 9: Integrate for Part (d)

For the condition \( y \leq x + 2 \), notice the bounding varies with \( x \). Split the area into two parts for \( 1 \leq x \leq 2 \) and below the line \( y=x+2 \): Solve \( \int_{3}^{5} \int_{1}^{y-2} f(x, y) \, dx \, dy \), confirming overlap and solve.

Step 10: Final Calculation for Part (d)

Calculate the integral using new bounds and check overlap in regions. Simplify and evaluate to find the probability.

Key Concepts

double integral

The double integral is a mathematical tool used to calculate the integral of a function over a two-dimensional area. When solving problems related to joint probability density functions, the double integral helps us assess the total probability over a specific region. In our given exercise, we perform a double integral over the rectangular region defined by the bounds:

• For the x-dimension: from 1 to 2
• For the y-dimension: from 3 to 5

First, we solve the inner integral with respect to one variable (in this case, x), keeping the other variable constant. Once we find this result, we perform the outer integral over the second variable, completing the double integral process. Step-by-step, the integration of each term in the function, followed by substitution of boundaries, gives us a result, which indicates the total probability density over the defined region. Verifying if this sum equals 1 confirms that the function qualifies as a joint probability density function over this area.

probability calculation

Calculating probability in the context of joint probability density functions involves setting up integrals over particular subregions of the defined area. Each subregion limits are based on specific constraints given in a problem. For instance, in our exercise, finding the probability for the dart landing between certain y-values requires integrating over reduced y-boundaries:

• For part (b): y is integrated from 3 to 4
• For part (c): y is integrated from 4 to 5

Again, the same double integral approach is followed, reducing the original bounds to these new limits. Calculating these probabilities reveals how likely events are, such as the dart landing in a particular area of the region. We calculate the double integral using these new limits, and each integration gives us a numerical probability that reflects the likelihood of the dart hitting a specified subregion within the full region.

probability density function

A probability density function (PDF) is critical to understand in statistics, especially when dealing with continuous variables. For joint distributions, the PDF determines how probability is distributed over continuous intervals of two variables. The given function in our exercise, defined as a combination of terms involving both x and y, illustrates this.

• The condition that the PDF must always be non-negative
• The integral of the function over the entire specified region must equal 1

This ensures the function appropriately models the behavior of the system or random event. In our exercise, the function given is a concrete example, and solving it within the context of the specified boundaries helps reinforce the concept that the PDF describes the "density" of probabilities across the range. Understanding this helps us analyze how different variables interact and what probabilities are assigned to values within given ranges.