/*! 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 19 A contractor is required by a co... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 19

A contractor is required by a county planning department to submit anywhere from one to five forms (depending on the nature of the project) in applying for a building permit. Let \(y\) be 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: \(\sum p(y)=1\) ) 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?

Short Answer

Expert verified
a. The value of \(k\) is 1/15. b. The probability that at most three forms are required is 0.4. c. The probability that between two and four forms are required is 0.6.

Step by step solution

01

Find the value of \(k\)

To find the value of \(k\), we need to remember that the total probability for all possible outcomes should add up to 1. So we get \(\sum_{y=1}^{5} k \times y = 1\). This becomes \(k \times (1 + 2 + 3 + 4 + 5) = 1\). This is a simple arithmetic sum which gives \(k \times 15 = 1\). So, \(k = 1/15\).
02

Find the probability that at most three forms are required

This requires summing up the probabilities for \(y = 1\), \(y = 2\) and \(y = 3\). So \(P(Y \leq 3) = k \times (1 + 2 + 3) = 1/15 \times 6 = 0.4\).
03

Find the probability that between two and four forms are required

This involves summing up the probabilities for \(y = 2\), \(y = 3\) and \(y = 4\). So \(P(2 \leq Y \leq 4) = k \times (2 + 3 + 4) = 1/15 \times 9 = 0.6\).

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 91影视!

Key Concepts

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

Probability Distribution
In probability theory, a probability distribution assigns probabilities to different possible outcomes of a random experiment. It essentially gives a complete description of the likelihood of various results happening.
Here's how it works:
  • A probability distribution is described using a mathematical function that lists all the possible outcomes and their corresponding probabilities.
  • The sum of all the probabilities in a probability distribution must equal 1, as this represents certainty that one of the outcomes will occur.
  • For example, in the problem given, the number of forms required is denoted by the variable \( y \), which ranges from 1 to 5 forms. The probability distribution in this scenario is given by \( p(y) = k \times y \), where \( k \) is a constant we solved for by ensuring that the total probability is 1.
Understanding probability distribution is key to making sense of how likely different outcomes are and calculating things like averages and variances. This concept is widely used in fields ranging from statistics to finance to science.
Discrete Random Variables
A discrete random variable is a type of variable used in statistics that can take on a countable number of distinct values. This is a concept essential to understanding scenarios where specific outcomes can be listed or counted.
Let's delve into the details:
  • Discrete random variables are often modeled by integers. For example, the number of forms required in the contractor problem is a discrete random variable, as it can be one of 1, 2, 3, 4, or 5 forms.
  • Each value the discrete random variable can take has an associated probability, as described by its probability distribution.
  • Unlike continuous random variables, which can take on any value within a range, discrete random variables have distinct, separated values.
Understanding discrete random variables helps in many practical situations. It's how we model scenarios where outcomes are specifically defined, like the roll of a dice or the number of times an event happens.
Mathematical Expectation
Mathematical expectation, also known as expected value, is a fundamental concept in probability that provides the average outcome of a random variable if we could repeat the experiment an infinite number of times. Although often associated with the mean in statistics, it's slightly different because it weighs each outcome by its probability.
The concept can be broken down as follows:
  • For a discrete random variable, the expected value is calculated by multiplying each possible outcome by its probability and summing all these products.
  • In our exercise, if we wanted to find the expected number of forms required, we would use the equation: \( E(y) = \sum_{y} y \times p(y) \).
  • This results in a single value that represents the 'center' or 'average' of the probability distribution, telling us what outcome we should expect on average, given the probabilities.
Mathematical expectation is utilized extensively to predict outcomes in a variety of fields. It's particularly useful in areas like finance and insurance, where understanding long-term averages is crucial.

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

Suppose that \(20 \%\) of all homeowners in an earthquakeprone area of California are insured against earthquake damage. Four homeowners are selected at random. Define the random variable \(x\) as the number among the four who have earthquake insurance. a. Find the probability distribution of \(x\). (Hint: Let \(S\) denote a homeowner who has insurance and \(\mathrm{F}\) one who does not. Then one possible outcome is SFSS, with probability (0.2)(0.8)(0.2)(0.2) and associated \(x\) value of 3 . There are 15 other outcomes.) b. What is the most likely value of \(x ?\) c. What is the probability that at least two of the four selected homeowners have earthquake insurance?

Of all airline flight requests received by a certain ticket broker, \(70 \%\) are for domestic travel \((\mathrm{D})\) and \(30 \%\) are for international flights (I). Define \(x\) to be the number that are for domestic flights among the next three requests received. Assuming independence of successive requests, determine the probability distribution of \(x\). (Hint: One possible outcome is DID, with the probability \((0.7)(0.3)(0.7)=0.147 .)\)

Suppose that \(20 \%\) of the 10,000 signatures on a certain recall petition are invalid. Would the number of invalid signatures in a sample of size 2,000 have (approximately) a binomial distribution? Explain.

According to the paper "Commuters' Exposure to Particulate Matter and Carbon Monoxide in Hanoi, Vietnam" (Transportation Research [2008]: 206-211), the carbon monoxide exposure of someone riding a motorbike for \(5 \mathrm{~km}\) on a highway in Hanoi is approximately normally distributed with a mean of 18.6 ppm. Suppose that the standard deviation of carbon monoxide exposure is 5.7 ppm. Approximately what proportion of those who ride a motorbike for \(5 \mathrm{~km}\) on a Hanoi highway will experience a carbon monoxide exposure of more than 20 ppm? More than \(25 \mathrm{ppm} ?\)

The time that it takes a randomly selected job applicant to perform a certain task has a distribution that can be approximated by a normal distribution with a mean of 120 seconds and a standard deviation of 20 seconds. The fastest \(10 \%\) are to be given advanced training. What task times qualify individuals for such training?
See all solutions

Recommended explanations on Math Textbooks

Discrete Mathematics

Read Explanation

Applied Mathematics

Read Explanation

Mechanics Maths

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Logic and Functions

Read Explanation
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.