/*! 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

The number of pumps in use at both a six-pump station and a four-pump station will be determined. Give the possible values for each of the following random variables: a. \(T=\) the total number of pumps in use b. \(X=\) the difference between the numbers in use at stations 1 and 2 c. \(U=\) the maximum number of pumps in use at either station d. \(Z=\) the number of stations having exactly two pumps in use

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The article "Reliability-Based Service-Life Assessment of Aging Concrete Structures" (J. Structural Engr., 1993: \(1600-1621\) ) suggests that a Poisson process can be used to represent the occurrence of structural loads over time. Suppose the mean time between occurrences of loads is .5 year. a. How many loads can be expected to occur during a 2year period? b. What is the probability that more than five loads occur during a 2-year period? c. How long must a time period be so that the probability of no loads occurring during that period is at most .1?

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Twenty percent of all telephones of a certain type are submitted for service while under warranty. Of these, \(60 \%\) can be repaired, whereas the other \(40 \%\) must be replaced with new units. If a company purchases ten of these telephones, what is the probability that exactly two will end up being replaced under warranty?
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