/*! 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] | 91Ó°ÊÓ

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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. 0.73; d. 0.53

Step by step solution

01

Understand the CDF

The cumulative distribution function (CDF), denoted as \(F(x)\), gives the probability that the random variable \(X\) takes on a value less than or equal to \(x\). For each piecewise component, \(F(x)\) accumulates the probability up to that \(x\) value.
02

Calculate P(X=2)

To find \(P(X=2)\), use the probability mass function, which is the difference in CDF values for adjacent intervals. \[ P(X=2) = F(3) - F(2) = 0.39 - 0.19 = 0.20 \]
03

Calculate P(X>3)

\(P(X>3)\) can be calculated using the complement rule: \[ P(X>3) = 1 - F(3) = 1 - 0.67 = 0.33 \]
04

Calculate P(2 ≤ X ≤ 5)

\(P(2 \leq X \leq 5)\) is the difference between the CDF at 5 and just before 2: \[ P(2 \leq X \leq 5) = F(5) - F(2) = 0.92 - 0.19 = 0.73 \]
05

Calculate P(2 < X < 5)

\(P(2 < X < 5)\) is the difference between the CDF at 5 and at 2: \[ P(2 < X < 5) = F(5) - F(3) = 0.92 - 0.39 = 0.53 \]

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Key Concepts

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

Probability Mass Function
The Probability Mass Function (PMF) is an essential concept when dealing with discrete random variables. For a random variable, the PMF gives the probability that the random variable takes on a specific value. It's crucial when working with cumulative distribution functions (CDFs) because it helps derive specific probabilities.When you read a CDF, it provides the cumulative probability up to a certain point. To find the probability of an exact value, such as \(P(X=2)\), you use the PMF by subtracting the CDF value just before from the value at that point:- Example: \(P(X=2) = F(3) - F(2) = 0.39 - 0.19 = 0.20\)In summary, the PMF helps drill down into the specific probabilities from the cumulative probabilities presented in CDFs. Therefore, mastering the use of the PMF is indispensable for solving discrete probability problems.
Piecewise Function
A piecewise function is a function that is defined by different expressions over different intervals of its input. In probability, piecewise functions often represent the cumulative distribution function (CDF), which gives us the cumulative probabilities across defined intervals.For example, in our given CDF:- For \(x < 0\), \(F(x) = 0\)- For \(0 \leq x < 1\), \(F(x) = 0.06\)- And so on...Each segment or 'piece' of this function defines the probability data in its range, allowing us to understand how probability accumulates across intervals.Piecewise functions are incredibly useful in probability calculations as they give precise probability metrics at different phases or checkpoints of the CDF, providing a clear breakdown of how probability aggregates through the variable's domain.
Complement Rule
The Complement Rule is a fundamental principle in probability theory used to simplify probability calculations. It states that the probability of an event not occurring is equal to one minus the probability of the event occurring. This rule is particularly useful when the direct calculation of a probability seems complex or when it is easier to calculate the probability of the opposite event. For example, to find \(P(X>3)\):- Instead of summing probabilities directly, use the complement of \(X \leq 3\):- So, \(P(X>3) = 1 - F(3) = 1 - 0.67 = 0.33\)The Complement Rule is not only a shortcut but also a crucial strategy for probability calculations, simplifying the computation process and helping you quickly find probabilities associated with the complementary events.
Probability Calculations
Probability Calculations are all about efficiently using mathematical tools to determine the likelihood of various outcomes. In our exercise, we had to calculate probabilities based on a CDF, such as \(P(2 \leq X \leq 5)\) and \(P(2 < X < 5)\).Steps usually involve:1. **Understanding the Question**: Break down what the probability asked for represents. Is it for exact values or a range?2. **Utilize CDF Data**: Use cumulative probabilities from a CDF. For ranges like \(P(2 \leq X \leq 5)\), subtract the CDF values at interval endpoints: - \(P(2 \leq X \leq 5) = F(5) - F(2) = 0.92 - 0.19 = 0.73\)3. **Adjust for Open Intervals**: For \(P(2 < X < 5)\), only adjust for endpoints: - \(P(2 < X < 5) = F(5) - F(3) = 0.92 - 0.39 = 0.53\)Following a systematic approach leads to clear, accurate probability calculations and ensures you account for all conditions specified in a probability problem.

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Most popular questions from this chapter

Suppose that you read through this year's issues of the New York Times and record each number that appears in a news article-the income of a CEO, the number of cases of wine produced by a winery, the total charitable contribution of a politician during the previous tax year, the age of a celebrity, and so on. Now focus on the leading digit of each number, which could be \(1,2, \ldots, 8\), or 9 . Your first thought might be that the leading digit \(X\) of a randomly selected number would be equally likely to be one of the nine possibilities (a discrete uniform distribution). However, much empirical evidence as well as some theoretical arguments suggest an alternative probability distribution called Benford's law: \(p(x)=P(1\) st digit is \(x)=\log _{10}\left(\frac{x+1}{x}\right) \quad x=1,2, \ldots, 9\) a. Without computing individual probabilities from this formula, show that it specifies a legitimate pmf. b. Now compute the individual probabilities and compare to the corresponding discrete uniform distribution. c. Obtain the cdf of \(X\). d. Using the cdf, what is the probability that the leading digit is at most 3 ? At least 5 ? [Note: Benford's law is the basis for some auditing procedures used to detect fraud in financial reporting-for example, by the Internal Revenue Service.]

After all students have left the classroom, a statistics professor notices that four copies of the text were left under desks. At the beginning of the next lecture, the professor distributes the four books in a completely random fashion to each of the four students \((1,2,3\), and 4 ) who claim to have left books. One possible outcome is that 1 receives 2 's book, 2 receives 4 's book, 3 receives his or her own book, and 4 receives l's book. This outcome can be abbreviated as \((2,4,3,1)\). a. List the other 23 possible outcomes. b. Let \(X\) denote the number of students who receive their own book. Determine the pmf of \(X\).

Let \(X\) be the number of material anomalies occurring in a particular region of an aircraft gas-turbine disk. The article "Methodology for Probabilistic Life Prediction of MultipleAnomaly Materials" (Amer. Inst. of Aeronautics and Astronautics \(J ., 2006: 787-793\) ) proposes a Poisson distribution for \(X\). Suppose that \(\mu=4\). a. Compute both \(P(X \leq 4)\) and \(P(X<4)\). b. Compute \(P(4 \leq X \leq 8)\). c. Compute \(P(8 \leq X)\). d. What is the probability that the number of anomalies exceeds its mean value by no more than one standard deviation?

Suppose small aircraft arrive at a certain airport according to a Poisson process with rate \(\alpha=8\) per hour, so that the number of arrivals during a time period of \(t\) hours is a Poisson rv with parameter \(\mu=8 t\). a. What is the probability that exactly 6 small aircraft arrive during a 1-hour period? At least 6? At least 10 ? b. What are the expected value and standard deviation of the number of small aircraft that arrive during a 90 -min period? c. What is the probability that at least 20 small aircraft arrive during a \(2.5\)-hour period? That at most 10 arrive during this period?

A small market orders copies of a certain magazine for its magazine rack each week. Let \(X=\) demand for the magazine, with pmf \begin{tabular}{l|llllll} \(x\) & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline\(p(x)\) & \(\frac{1}{15}\) & \(\frac{2}{15}\) & \(\frac{3}{15}\) & \(\frac{4}{15}\) & \(\frac{3}{15}\) & \(\frac{2}{15}\) \end{tabular} Suppose the store owner actually pays \(\$ 2.00\) for each copy of the magazine and the price to customers is \(\$ 4.00\). If magazines left at the end of the week have no salvage value, is it better to order three or four copies of the magazine? [Hint: For both three and four copies ordered, express net revenue as a function of demand \(X\), and then compute the expected revenue.]

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