/*! 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 23 A consumer organization that eva... [FREE SOLUTION] | 91影视

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Chapter 3: Problem 23

A consumer organization that evaluates new automobiles customarily reports the number of major defects in each car examined. Let \(X\) denote the number of major defects in a randomly selected car of a certain type. The cdf of \(X\) is as follows: $$ F(x)= \begin{cases}0 & x<0 \\ .06 & 0 \leq x<1 \\ .19 & 1 \leq x<2 \\ .39 & 2 \leq x<3 \\ .67 & 3 \leq x<4 \\ .92 & 4 \leq x<5 \\ .97 & 5 \leq x<6 \\\ 1 & 6 \leq x\end{cases} $$ Calculate the following probabilities directly from the cdf: a. \(p(2)\), that is, \(P(X=2)\) b. \(P(X>3)\) c. \(P(2 \leq X \leq 5)\) d. \(P(2

Short Answer

Expert verified
a. 0.20, b. 0.33, c. 0.58, d. 0.53

Step by step solution

01

Calculate P(X=2)

The probability of a discrete random variable taking a specific value is the difference between the cumulative distribution function (CDF) values at that value and the previous integer. Therefore, the probability that a car has exactly 2 major defects, \( P(X=2) \), is calculated by finding \( F(2) - F(1) \).From the given CDF, \( F(2) = 0.39 \) and \( F(1) = 0.19 \).So, \[ P(X=2) = F(2) - F(1) = 0.39 - 0.19 = 0.20 \].
02

Calculate P(X>3)

To find the probability that a random variable is greater than a certain value, we subtract the CDF value at that point from 1. Here, we want \( P(X>3) \).We know \( F(3) = 0.67 \), so:\[ P(X > 3) = 1 - F(3) = 1 - 0.67 = 0.33 \]
03

Calculate P(2 鈮� X 鈮� 5)

The probability that \( X \) lies in an interval \( [a, b] \) is given by the difference \( F(b+1) - F(a) \). Here, we need \( P(2 \leq X \leq 5) \).Since \( F(5) = 0.97 \) and \( F(2) = 0.39 \), we calculate:\[ P(2 \leq X \leq 5) = F(5+1) - F(2) = F(6) - F(2) = 1 - 0.39 = 0.58 \]
04

Calculate P(2 < X < 5)

For the probability of \( X \) between non-inclusive boundaries \( (a, b) \), we use \( F(b) - F(a+1) \). Therefore, the probability \( P(2 < X < 5) \) is found as follows.Given \( F(4) = 0.92 \) and \( F(3) = 0.67 \):\[ P(2 < X < 5) = F(5) - F(3) = 0.92 - 0.39 = 0.53 \]

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Key Concepts

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

Probability Calculations
Understanding probability calculations is key to grasping how we analyze the likelihood of different outcomes of a random variable. When given the cumulative distribution function (CDF), like in our exercise, we can extract probabilities by focusing on the difference in values between specific points on the CDF.

Let's explore this further:
  • For a discrete random variable, the probability of it taking a specific value, say at point 2, is calculated by finding the difference in the CDF values at that point and the previous integer. This is shown in the calculation of \( P(X=2) \) where \( F(2) - F(1) \) gives us the probability of 0.20.
  • To calculate the probability of \( X \) being greater than a certain value, say 3, you subtract the CDF value at that point from 1. Therefore, \( P(X>3) \) is calculated as \( 1 - F(3) \), resulting in 0.33.
  • Probabilities for intervals, like \( P(2 \leq X \leq 5) \), use the difference between distributing values at the endpoints, resulting in 0.58 in our example.
These techniques provide a systematic way to deduce probabilities from a CDF, enhancing our understanding of the behavior of discrete random variables.
Discrete Random Variable
A discrete random variable, such as the one in this exercise, is a variable that can take on a finite or countably infinite number of distinct values. In our context, it represents the number of major defects in a car. Each potential outcome, or number of defects, is associated with a probability derived from the CDF.

Here鈥檚 what you need to understand about discrete random variables:
  • The outcomes are countable. For instance, a car can have 0, 1, 2, etc., defects, but not 1.5 defects.
  • Each specific outcome has a specific probability, calculated by finding the change in CDF values at key integer points.
  • The sum of the probabilities over all possible outcomes is always 1.
In our exercise, the values of the CDF at specific points reveal the probabilities of a car having a precise number of defects. This helps in quality studies and predictions.
Probability Distribution
A probability distribution describes how the probabilities are distributed over the values that the random variable can take. For a discrete random variable like in our exercise, the CDF provides a way to map each potential outcome to its cumulative probability.

Key points about probability distributions include:
  • The cumulative distribution function (CDF) aids in evaluating the likelihood that a random variable will assume a value less than or equal to a certain level.
  • With it, we can determine any individual or range of outcomes' probabilities. For instance, using the CDF to find \( P(2 < X < 5) \) as 0.53.
  • Knowing how to translate CDF values into probability insights involves understanding that the steeper the inclination of the function at any point, the more concentrated the probability mass.
Understanding this distribution allows analysts to make informed predictions and decisions based on potential occurrences.

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Most popular questions from this chapter

For each random variable defined here, describe the set of possible values for the variable, and state whether the variable is discrete. a. \(X=\) the number of unbroken eggs in a randomly chosen standard egg carton b. \(Y=\) the number of students on a class list for a particular course who are absent on the first day of classes c. \(U=\) the number of times a duffer has to swing at a golf ball before hitting it d. \(X=\) the length of a randomly selected rattlesnake e. \(Z=\) the amount of royalties earned from the sale of a first edition of 10,000 textbooks f. \(Y=\) the pH of a randomly chosen soil sample g. \(X=\) the tension (psi) at which a randomly selected tennis racket has been strung h. \(X=\) the total number of coin tosses required for three individuals to obtain a match (HHH or TTT)

Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\lambda=20\) (suggested in the article "Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20 ? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

a. Show that \(b(x ; n, 1-p)=b(n-x ; n, p)\). b. Show that \(B(x ; n, 1-p)=1-B(n-x-1 ; n, p)\). [Hint: At most \(x S\) 's is equivalent to at least \((n-x) F\) 's.] c. What do parts (a) and (b) imply about the necessity of including values of \(p\) greater than \(.5\) in Appendix Table A.1?

Suppose that \(90 \%\) of all batteries from a certain supplier have acceptable voltages. A certain type of flashlight requires two type-D batteries, and the flashlight will work only if both its batteries have acceptable voltages. Among ten randomly selected flashlights, what is the probability that at least nine will work? What assumptions did you make in the course of answering the question posed?

An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let \(X=\) the number of months between successive payments. The cdf of \(X\) is as follows: $$ F(x)= \begin{cases}0 & x<1 \\ .30 & 1 \leq x<3 \\ .40 & 3 \leq x<4 \\ .45 & 4 \leq x<6 \\ .60 & 6 \leq x<12 \\ 1 & 12 \leq x\end{cases} $$ a. What is the pmf of \(X\) ? b. Using just the cdf, compute \(P(3 \leq X \leq 6)\) and \(P(4 \leq X)\).
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