/*! 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 14 A contractor is required by a co... [FREE SOLUTION] | 91Ó°ÊÓ

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A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature of the project) in applying for a building permit. Let \(Y=\) the number of forms required of the next applicant. The probability that \(y\) forms are required is known to be proportional to \(y\)-that is, \(p(y)=k y\) for \(y=1, \ldots, 5\). a. What is the value of \(k\) ? [Hint: \(\left.\sum_{y=1}^{5} p(y)=1 .\right]\) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required? d. Could \(p(y)=y^{2} / 50\) for \(y=1, \ldots, 5\) be the pmf of \(Y\) ?

Short Answer

Expert verified
a. \(k=\frac{1}{15}\). b. \(\frac{2}{5}\). c. \(\frac{3}{5}\). d. No, it is not a valid PMF.

Step by step solution

01

Set up the Probability Mass Function (PMF)

Given that the probability function is proportional to the number of forms, start by expressing it as: \( p(y) = k \cdot y \) for \( y = 1, 2, 3, 4, 5 \).
02

Use the Total Probability Theorem to Find k

Use the condition that probabilities sum to 1: \[ \sum_{y=1}^{5} p(y) = 1. \] Substituting the expression for \( p(y) \), we have: \( k \cdot 1 + k \cdot 2 + k \cdot 3 + k \cdot 4 + k \cdot 5 = 1 \). Simplify to obtain \( k \cdot (1 + 2 + 3 + 4 + 5) = 1 \), which simplifies to \( k \cdot 15 = 1 \). Solve for \( k \) to get \( k = \frac{1}{15} \).
03

Calculate Probability that at Most Three Forms are Required

Sum the probabilities for 1, 2, and 3 forms: \[ P(Y \leq 3) = p(1) + p(2) + p(3) \] \[ = \left( \frac{1}{15} \cdot 1 \right) + \left( \frac{1}{15} \cdot 2 \right) + \left( \frac{1}{15} \cdot 3 \right) = \frac{1}{15} + \frac{2}{15} + \frac{3}{15} = \frac{6}{15} = \frac{2}{5}. \]
04

Calculate Probability that Between Two and Four Forms are Required

Calculate the probability for 2, 3, and 4 forms: \[ P(2 \leq Y \leq 4) = p(2) + p(3) + p(4) \] \[ = \left( \frac{1}{15} \cdot 2 \right) + \left( \frac{1}{15} \cdot 3 \right) + \left( \frac{1}{15} \cdot 4 \right) = \frac{2}{15} + \frac{3}{15} + \frac{4}{15} = \frac{9}{15} = \frac{3}{5}. \]
05

Verify the Validity of Given PMF

Check if \( p(y) = \frac{y^2}{50} \) can be a valid PMF. Compute \( \sum_{y=1}^{5} \frac{y^2}{50} \): \[ \frac{1^2}{50} + \frac{2^2}{50} + \frac{3^2}{50} + \frac{4^2}{50} + \frac{5^2}{50} = \frac{1}{50} + \frac{4}{50} + \frac{9}{50} + \frac{16}{50} + \frac{25}{50} = \frac{55}{50} = 1.1. \] Since the sum is greater than 1, \( p(y) = \frac{y^2}{50} \) is not a valid PMF.

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Key Concepts

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

Probability Theory
Probability theory is the branch of mathematics that deals with the likelihood of events occurring. In the context of the exercise, we're dealing with discrete events, which are specific occurrences that have a fixed set of possible outcomes. Probabilities are assigned to these outcomes to quantify how likely they are to happen.

The concept of a probability mass function (PMF) is at the heart of this. A PMF maps each possible outcome, like the number of forms in the exercise, to a probability. However, the probabilities of all these possible outcomes must sum up to 1. This ensures that one of the outcomes definitely occurs. For example, in our case:
  • The PMF is given by the formula: \(p(y) = ky\), for \(y = 1, 2, 3, 4, 5\).
  • The total probability is expressed as \(\sum_{y=1}^{5} p(y) = 1\).
  • This equation allows us to determine the proportionality constant \(k\), which is crucial for balancing the PMF.
Thus, probability theory provides the mathematical backbone for determining how likely different numbers of forms are required in this context.
Discrete Random Variables
In probability theory, a random variable is a variable that has a random value that is subject to variations due to chance. Specifically, a discrete random variable is one that can take on only a countable number of distinct values.

In the exercise, the random variable \(Y\) represents the number of forms required, and it can only be integers from 1 to 5. Each possible outcome or value of \(Y\) is associated with a probability through the probability mass function (PMF).

Key features of discrete random variables include:
  • The set of possible outcomes is countable and finite—meaning we can list them one by one, such as 1, 2, 3, 4, and 5 forms in this problem.
  • The probability for each outcome depends on a defined PMF, like \(p(y) = \/(k \cdot y\) in our problem.
  • These probabilities offer insight into the distribution of outcomes for the random variable \(Y\).
Understanding discrete random variables allows us to tackle problems by calculating probabilities for specific ranges, such as finding the probability that at most three forms are required or that between two to four forms are necessary.
Statistical Analysis
Statistical analysis involves collecting, exploring, and interpreting complex data, and it often uses probability theory to derive meaningful insights. In the current exercise, we're conducting a statistical analysis on the number of building permit forms required.

A crucial part of such analysis is determining the probabilities for different scenarios based on the given PMF. We solve problems like:
  • Calculating the probability of at most three forms being required, which involves summing up the probabilities for each outcome—1, 2, and 3 forms.
  • Determining the probability for having forms between 2 and 4, inclusive, which requires adding the probabilities for outcomes 2, 3, and 4.
  • Verifying the validity of a PMF, as shown in the exercise, by checking if an alternative PMF could theoretically sum to one.
Having such proficiency in statistical analysis not only aids in tackling specific problems but also enhances our overall ability to work with data more effectively and intuitively.

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Most popular questions from this chapter

A family decides to have children until it has three children of the same gender. Assuming \(P(B)=P(G)=.5\), what is the pmf of \(X=\) the number of children in the family?

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