/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 Let \(Y_{1}\) and \(Y_{2}\) deno... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(Y_{1}\) and \(Y_{2}\) denote the proportions of time (out of one workday) during which employees I and II, respectively, perform their assigned tasks. The joint relative frequency behavior of \(Y_{1}\) and \(Y_{2}\) is modeled by the density function $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} y_{1}+y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find \(P\left(Y_{1}<1 / 2, Y_{2}>1 / 4\right)\). b. Find \(P\left(Y_{1}+Y_{2} \leq 1\right)\).

Short Answer

Expert verified
a. \(\frac{21}{64}\) b. \(\frac{1}{3}\)

Step by step solution

01

Define the Region of Integration for Part (a)

We need to find the probability that \(Y_1 < \frac{1}{2}\) and \(Y_2 > \frac{1}{4}\). This translates to finding the area in the region of the joint plane \(0 \leq y_1 \leq \frac{1}{2}\) and \(\frac{1}{4} < y_2 \leq 1\). The area of integration is from \(0\) to \(\frac{1}{2}\) for \(y_1\) and \(\frac{1}{4}\) to \(1\) for \(y_2\).
02

Set Up the Integral for Part (a)

The probability is given by integrating the joint density function over the region defined in Step 1. Thus, the integral is: \[ P(Y_1 < \frac{1}{2}, Y_2 > \frac{1}{4}) = \int_{0}^{1/2} \int_{1/4}^{1} (y_1 + y_2) \, dy_2 \, dy_1 \]
03

Integrate with Respect to \(y_2\) for Part (a)

First, integrate the function \(y_1 + y_2\) with respect to \(y_2\): \[ \int_{1/4}^{1} (y_1 + y_2) \, dy_2 = \left[ y_1y_2 + \frac{y_2^2}{2} \right]_{1/4}^{1} \]\[ = \left( y_1 \cdot 1 + \frac{1^2}{2} \right) - \left( y_1 \cdot \frac{1}{4} + \frac{(1/4)^2}{2} \right) \]\[ = (y_1 + \frac{1}{2}) - (\frac{y_1}{4} + \frac{1}{32}) \]
04

Simplify and Integrate with Respect to \(y_1\) for Part (a)

Simplify the result of the integration: \[ = \frac{3y_1}{4} + \frac{15}{32} \]Then integrate over \(y_1\) from \(0\) to \(\frac{1}{2}\): \[ \int_{0}^{1/2} \left( \frac{3y_1}{4} + \frac{15}{32} \right) \, dy_1 \]\[ = \left[ \frac{3y_1^2}{8} + \frac{15y_1}{32} \right]_{0}^{1/2} \]\[ = \left( \frac{3(1/2)^2}{8} + \frac{15(1/2)}{32} \right) - 0 \]\[ = \frac{3}{32} + \frac{15}{64} = \frac{21}{64} \]
05

Verify Solution for Part (a)

Ensure all work is correct. The probability calculated is indeed \( P(Y_1 < 1/2, Y_2 > 1/4) = \frac{21}{64} \).
06

Define the Region of Integration for Part (b)

Find the probability that \(Y_1 + Y_2 \leq 1\). The region is the area under the line \(y_1 + y_2 = 1\) within the valid range for \(y_1\) and \(y_2\), which is a triangle with vertices at \((0, 0)\), \((1, 0)\), and \((0, 1)\).
07

Set Up the Integral for Part (b)

The integral will be evaluated over the triangular region: \[ P(Y_1 + Y_2 \leq 1) = \int_{0}^{1} \int_{0}^{1-y_1} (y_1 + y_2) \, dy_2 \, dy_1 \]
08

Integrate with Respect to \(y_2\) for Part (b)

First, integrate the function \(y_1 + y_2\) with respect to \(y_2\) from \(0\) to \(1-y_1\): \[ \int_{0}^{1-y_1} (y_1 + y_2) \, dy_2 = \left[ y_1y_2 + \frac{y_2^2}{2} \right]_{0}^{1-y_1} \]\[ = y_1(1-y_1) + \frac{(1-y_1)^2}{2} \]\[ = y_1 - y_1^2 + \frac{1}{2} - y_1 + \frac{y_1^2}{2} \]\[ = -\frac{y_1^2}{2} + \frac{1}{2} \]
09

Integrate with Respect to \(y_1\) for Part (b)

Now integrate with respect to \(y_1\) from \(0\) to \(1\): \[ \int_{0}^{1} \left(-\frac{y_1^2}{2} + \frac{1}{2} \right) \, dy_1 \]\[ = \left[ -\frac{y_1^3}{6} + \frac{y_1}{2} \right]_{0}^{1} \]\[ = \left(-\frac{1}{6} + \frac{1}{2}\right) - 0 \]\[ = \frac{1}{3} \]
10

Verify Solution for Part (b)

Check the calculations; ensure the integration and substitutions are correct. The probability is \( P(Y_1 + Y_2 \leq 1) = \frac{1}{3} \).

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Key Concepts

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

Density Function
A density function, in the context of continuous probability distributions, helps determine the likelihood of a random variable taking on a particular range of values. It is analogous to a histogram in a discrete setting. For continuous variables, instead of counting occurrences, we integrate over intervals. This function must satisfy two key properties:
  • It is non-negative for all values in its domain.
  • The integral of the density function over the entire space must equal 1, ensuring total probability.
In our example, the density function is given by \(f(y_1, y_2) = y_1 + y_2\) for \(0 \leq y_1 \leq 1\) and \(0 \leq y_2 \leq 1\). This specific function models how employees I and II spend their workday. If either of these proportions veers outside 0 to 1, the function returns 0, capturing the infeasibility of such outcomes.

Understanding the density function is crucial, as it helps us set up and evaluate integrals for probabilities within specified regions.
Integration
Integration is the mathematical process of finding the area under a curve described by a mathematical function. In the context of probability, integration plays a pivotal role in determining the likelihood of a variable falling within a range. It transforms the density function, a representation of rate, into the probability, a tangible measure.

When computing probabilities for continuous variables, we integrate over the region of interest. For instance, in Part (a) of the problem, we aim to determine the probability that \(Y_1 < \frac{1}{2}\) and \(Y_2 > \frac{1}{4}\). The corresponding integral was set up as:
  • First, integrate \(y_1 + y_2\) with respect to \(y_2\).
  • Then, integrate the resulting expression with respect to \(y_1\).
The process involves determining limits of integration based on intersections of constraints and the potential outcomes permitted by the density function. Evaluating each integral carefully ensures an accurate probability measure. Integration helps bridge our understanding from abstract density functions to concrete probability values.
Probability Calculation
Calculating probabilities with respect to a joint density function involves determining the likelihood of events occurring simultaneously for two or more random variables. The joint probability is obtained by integrating the joint density function over the specified region.

Consider Part (a) of our problem, where we calculated \(P(Y_1 < \frac{1}{2}, Y_2 > \frac{1}{4})\). Here, the joint density function \(f(y_1, y_2)\) was integrated over this specific region yielding:
\[ \int_{0}^{1/2} \int_{1/4}^{1} (y_1 + y_2) \, dy_2 \, dy_1 = \frac{21}{64}\] The calculated value \(\frac{21}{64}\) represents the proportion of total probability mass accountable by the condition \(Y_1 < \frac{1}{2}\) and \(Y_2 > \frac{1}{4}\). Evaluating regions of integration correctly according to conditions is vital for legitimate probability estimations.
Joint Density Function
A joint density function describes the probabilistic behavior of two or more random variables considered together. It gives insights into how the variables might influence each other and the likelihood of their respective outcomes coinciding.
  • The joint density function is denoted as \(f(y_1, y_2)\) in this context, characterizing how much time employees spend on tasks.
  • It captures the co-dependencies between \(Y_1\) and \(Y_2\) by assigning a joint probability to intervals defined over both variables.
In Part (b), the probability \(P(Y_1 + Y_2 \leq 1)\) was determined through the function's incorporation. The integration was evaluated over a triangular region yielding \(\frac{1}{3}\). This value indicates the shared probability captured by the constraint \(Y_1 + Y_2 \leq 1\). Proper comprehension of the joint density function facilitates accurate probabilistic reasoning about multi-variable scenarios, catering to more nuanced interactions of variables beyond solo variable configurations.

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Most popular questions from this chapter

In Exercise \(5.12,\) we were given the following joint probability density function for the random variables \(Y_{1}\) and \(Y_{2},\) which were the proportions of two components in a sample from a mixture of insecticide: $$ f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 2, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1,0 \leq y_{1}+y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right. $$ For the two chemicals under consideration, an important quantity is the total proportion \(Y_{1}+Y_{2}\) found in any sample. Find \(E\left(Y_{1}+Y_{2}\right)\) and \(V\left(Y_{1}+Y_{2}\right)\)

A technician starts a job at a time \(Y_{1}\) that is uniformly distributed between 8: 00 A.M. and 8: 15 A.M. The amount of time to complete the job, \(Y_{2}\), is an independent random variable that is uniformly distributed between 20 and 30 minutes. What is the probability that the job will be completed before 8:30 A.M.?

Let \(Y_{1}\) and \(Y_{2}\) be independent exponentially distributed random variables, each with mean 1 . Find \(P\left(Y_{1}>Y_{2} | Y_{1}<2 Y_{2}\right)\).

In Exercise \(5.18, Y_{1}\) and \(Y_{2}\) denoted the lengths of life, in hundreds of hours, for components of types I and II, respectively, in an electronic system. The joint density of \(Y_{1}\) and \(Y_{2}\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} (1 / 8) y_{1} e^{-\left(y_{1}+y_{2}\right) / 2}, & y_{1} > 0, y_{2} > 0 \\ 0, & \text { elsewhere } \end{array}\right.$$

Let \(Y_{1}\) and \(Y_{2}\) have the joint probability density function given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} k y_{1} y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1, \\ 0, & \text { elsewhere } \end{array}\right.$$ a. Find the value of \(k\) that makes this a probability density function. b. Find the joint distribution function for \(Y_{1}\) and \(Y_{2}\). c. Find \(P\left(Y_{1} \leq 1 / 2, Y_{2} \leq 3 / 4\right)\).

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