/*! 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 4 A small-business Web site contai... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A small-business Web site contains 100 pages and \(60 \%, 30 \%,\) and \(10 \%\) of the pages contain low, moderate, and high graphic content, respectively. A sample of four pages is selected randomly without replacement, and \(X\) and \(Y\) denote the number of pages in the sample with moderate and high graphics output. Determine: a. \(f_{X Y}(x, y)\) b. \(f_{X}(x)\) c. \(E(X)\)

Short Answer

Expert verified
a. Use hypergeometric distribution for joint probability. b. Summation over \(y\) gives marginal distribution. c. \(E(X) = 1.2\).

Step by step solution

01

Understand the Problem

We have a web page consisting of 100 pages. Of these, 60% have low graphics, 30% have moderate graphics, and 10% have high graphics. We are sampling four pages without replacement, and need to find the joint distribution of moderate (\(X\)) and high graphics (\(Y\)) pages.
02

Define the Random Variables

Let \(X\) be the number of pages in the sample that have moderate graphics. Since moderate graphics make up 30% of the site or 30 pages, \(X\) follows a hypergeometric distribution with parameters \(N=100\), \(K=30\), and \(n=4\). Similarly, let \(Y\) be the number of pages with high graphics. High graphics make up 10% or 10 pages, so \(Y\) also follows a hypergeometric distribution with parameters \(N=100\), \(K=10\), and \(n=4\).
03

Joint Probability Distribution \(f_{X Y}(x, y)\)

The joint probability \(f_{XY}(x, y)\) is given by the hypergeometric distribution for both \(x\) moderate and \(y\) high graphics pages, considering that the sum of moderate, high and the remaining pages must be 4. It's given by:\[ f_{X Y}(x, y) = \frac{\binom{30}{x} \binom{10}{y} \binom{60}{4-x-y}}{\binom{100}{4}} \]where \(x + y \leq 4\) and \(x, y \geq 0\).
04

Marginal Probability Distribution \(f_{X}(x)\)

To find \(f_{X}(x)\), sum the joint probability over all possible values of \(y\):\[ f_{X}(x) = \sum_{y=0}^{4-x} f_{X Y}(x, y) \]Compute this using the equation from Step 3 and considering all possible valid \(y\) for each \(x\) given the sample size.
05

Expectation \(E(X)\)

The expected value of \(X\) is given by the formula for expectation in hypergeometric distributions. Calculate as follows:\[ E(X) = n \times \frac{K}{N} = 4 \times \frac{30}{100} = 1.2 \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Joint Probability Distribution
In probability theory, a joint probability distribution describes the likelihood of two or more random variables occurring simultaneously. For this exercise, we are interested in finding the joint probability distribution of two variables: \(X\) and \(Y\), where \(X\) represents the number of sampled pages with moderate graphics, and \(Y\) represents those with high graphics.
Here, the joint probability distribution \(f_{XY}(x, y)\) is derived using the hypergeometric distribution. This distribution is suitable because we are sampling without replacement from a finite population. The population consists of 100 web pages, out of which 30 have moderate graphics and 10 have high graphics.
The joint probability function is given by:
  • \[ f_{XY}(x, y) = \frac{\binom{30}{x} \binom{10}{y} \binom{60}{4-x-y}}{\binom{100}{4}} \]
This formula calculates the probability of selecting \(x\) pages with moderate graphics and \(y\) with high graphics, ensuring that their sum along with any pages with low graphics equals our total sample size of 4. The variables \(x\) and \(y\) must satisfy \(x + y \leq 4\) and \(x, y \geq 0\).
This joint distribution reflects all combinations of moderate and high graphics possibilities in the sample, providing a comprehensive view of their probabilities.
Marginal Probability Distribution
Marginal probability distribution is the probability distribution of a subset of variables within a larger set when the remaining variables are not specified or considered. In this exercise, we focus on finding the marginal probability distribution of the variable \(X\), the number of pages with moderate graphics.
To derive the marginal probability distribution \(f_X(x)\), we sum the joint probability distribution over all possible values of \(Y\). This involves considering all feasible values of \(y\) that can be paired with \(x\) such that their sum does not exceed the sample size. For each specific \(x\), we calculate:
  • \[ f_X(x) = \sum_{y=0}^{4-x} f_{XY}(x, y) \]
By performing this summation, we extract the probabilities associated solely with \(x\), the moderate graphics pages, effectively "marginalizing" over \(y\).
This marginal distribution helps us understand the likelihood of observing a given number of moderate graphics pages within our sample, irrespective of the number of high graphics pages included.
Expectation in Probability
Expectation in the context of probability, often referred to as the expected value or mean, represents the average outcome of a random variable if the experiment were repeated many times. For a hypergeometric distribution, which applies to our scenario of sampling without replacement, the expected value can be calculated using a simple formula.
For our variable \(X\), the expected number of pages with moderate graphics is determined with:
  • \[ E(X) = n \times \frac{K}{N} = 4 \times \frac{30}{100} = 1.2 \]
Here,
  • \(n\) is the sample size, which is 4.
  • \(K\) is the number of pages with moderate graphics in the entire website, which is 30.
  • \(N\) is the total number of pages, which is 100.
Thus, the expected value \(E(X)\) equals 1.2, indicating that on average, one can anticipate selecting about 1.2 pages with moderate graphics in a random sample of four pages.
Expectation gives us an idea of the central tendency of a probability distribution and is a crucial concept in both theoretical and applied probability.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

US In the manufacture of electroluminescent lamps, several different layers of ink are deposited onto a plastic substrate. The thickness of these layers is critical if specifications regarding the final color and intensity of light are to be met. Let \(X\) and \(Y\) denote the thickness of two different layers of ink. It is known that \(X\) is normally distributed with a mean of 0.1 millimeter and a standard deviation of 0.00031 millimeter, and \(Y\) is normally distributed with a mean of 0.23 millimeter and a standard deviation of 0.00017 millimeter. The value of \(\rho\) for these variables is equal to \(0 .\) Specifications call for a lamp to have a thickness of the ink corresponding to \(X\) in the range of 0.099535 to 0.100465 millimeter and \(Y\) in the range of 0.22966 to 0.23034 millimeter. What is the probability that a randomly selected lamp will conform to specifications?

The power in a DC circuit is \(P=I^{2} / R\) where \(I\) and \(R\) denote the current and resistance, respectively. Suppose that \(I\) is approximately normally distributed with mean of \(200 \mathrm{~mA}\) and standard deviation \(0.2 \mathrm{~mA}\) and \(R\) is a constant. Determine the probability density function of power.

The intensity \(\left(\mathrm{mW} / \mathrm{mm}^{2}\right)\) of a laser beam on a surface theoretically follows a bivariate normal distribution with maximum intensity at the center, equal variance \(\sigma\) in the \(x\) and \(y\) directions, and zero covariance. There are several definitions for the width of the beam. One definition is the diameter at which the intensity is \(50 \%\) of its peak. Suppose that the beam width is \(1.6 \mathrm{~mm}\) under this definition. Determine \(\sigma\). Also determine the beam width when it is defined as the diameter where the intensity equals \(1 / e^{2}\) of the peak.

Suppose that \(X\) and \(Y\) have a bivariate normal distribution with \(\sigma_{X}=0.04, \sigma_{Y}=0.08, \mu_{X}=3.00, \mu_{Y}=7.70,\) and \(\rho=0\). Determine the following: a. \(P(2.95

Let \(X\) and \(Y\) represent the concentration and viscosity of a chemical product. Suppose that \(X\) and \(Y\) have a bivariate normal distribution with \(\sigma_{X}=4, \sigma_{Y}=1, \mu_{X}=2,\) and \(\mu_{Y}=1 .\) Draw a rough contour plot of the joint probability density function for each of the following values of \(\rho\) : a. \(\rho=0\) b. \(\rho=0.8\) c. \( \rho=-0.8\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.