/*! 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 16 Let \(X\) be a continuous rv wit... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(X\) be a continuous rv with cdf $$ F(x)=\left\\{\begin{array}{cl} 0 & x \leq 0 \\ \frac{x}{4}\left[1+\ln \left(\frac{4}{x}\right)\right] & 04 \end{array}\right. $$ [This type of cdf is suggested in the article "Variability in Measured Bedload-Transport Rates" (Water 91Ó°ÊÓ Bull., 1985:39-48) as a model for a hydrologic variable.] What is a. \(P(X \leq 1)\) ? b. \(P(1 \leq X \leq 3)\) ? c. The pdf of \(X\) ?

Short Answer

Expert verified
a. 0.6 \n b. 0.263 \n c. \(f(x) = \frac{1}{4}\left[1+\ln\left(\frac{4}{x}\right) - \frac{1}{x}\right]\) for \(0 < x \leq 4\)

Step by step solution

01

Understanding the CDF

We have a cumulative distribution function (CDF) given piecewise. This function, denoted as \(F(x)\), describes the probability that the random variable \(X\) is less than or equal to \(x\). The function is defined as follows: - \(F(x) = 0\) for \(x \leq 0\).- \(F(x) = \frac{x}{4}\left[1+\ln \left(\frac{4}{x}\right)\right]\) for \(0 < x \leq 4\).- \(F(x) = 1\) for \(x > 4\).We use this information to find probabilities and derive the CDF.
02

Calculate P(X ≤ 1)

Given the definition of the CDF, \(P(X \leq 1)\) is found directly using \(F(1)\). For \(0 < x \leq 4\), the CDF is given by \(F(x) = \frac{x}{4}\left[1 + \ln\left(\frac{4}{x}\right)\right]\). Substitute \(x = 1\):\[ F(1) = \frac{1}{4}\left[1 + \ln\left(\frac{4}{1}\right)\right] = \frac{1}{4}(1 + \ln(4)) \approx 0.6 \]
03

Calculate P(1 ≤ X ≤ 3)

To find this probability, use the CDF evaluated at different limits: \(P(1 \leq X \leq 3) = F(3) - F(1)\). First, calculate \(F(3)\):\[ F(3) = \frac{3}{4}\left[1 + \ln\left(\frac{4}{3}\right)\right] \approx 0.863 \]Using our previous result for \(F(1) \approx 0.6\), we compute:\[ P(1 \leq X \leq 3) = 0.863 - 0.6 = 0.263 \]
04

Derive the PDF from the CDF

The probability density function (PDF) \(f(x)\) is the derivative of the CDF \(F(x)\). For \(0 < x \leq 4\), we differentiate \(F(x)\):\[ f(x) = \frac{d}{dx}\left[\frac{x}{4}\left(1 + \ln\left(\frac{4}{x}\right)\right)\right] \]Apply the product and chain rules:\[ f(x) = \frac{1}{4}\left[1 + \ln \left(\frac{4}{x}\right)\right] - \frac{1}{4x} \]Thus, the PDF is \(f(x) = \frac{1}{4}\left[1+\ln\left(\frac{4}{x}\right) - \frac{1}{x}\right]\) for \(0 < x \leq 4\).

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Key Concepts

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

Continuous Random Variable (rv)
A continuous random variable, often abbreviated as rv, is a variable that can take an infinite number of different values within a given range. This range is typically over the set of real numbers, meaning the variable can take any value within a specified interval. For example, if we are looking at temperature over a day, it can vary continuously throughout the time. When dealing with continuous random variables, we often define their behavior using a Cumulative Distribution Function (CDF). The CDF gives the probability that the random variable will take a value less than or equal to a particular number. In our exercise, we have a random variable, denoted as \(X\), with a specified CDF valid for different ranges. Understanding the behavior of \(X\) includes recognizing when the CDF assumes different values like zero or one, depending on the intervals.
Probability Density Function (PDF)
The Probability Density Function (PDF) is crucial when working with continuous random variables. It tells us how the probability is distributed across different values of the variable. Essentially, the PDF describes the likelihood of the random variable taking on a specific value.In the context of our exercise, the PDF is derived from the given CDF. To find the PDF, we differentiate the CDF with respect to \(x\). This differentiation gives us a function describing the probability distribution over the continuous range. For our particular CDF, the PDF formula is determined by calculating the derivative of \(F(x)\) for \(0 < x \leq 4\), which results in:\[ f(x) = \frac{1}{4}\left[1+\ln\left(\frac{4}{x}\right) - \frac{1}{x}\right] \]This function gives us a detailed view of how probable each value within this interval is. It is important to remember that while the PDF provides insight into how data points are spread, its value is not a direct probability.
Probability Calculations
Performing probability calculations for continuous random variables often involves working with the CDF and PDF. In the step-by-step solution, we calculated the probability that \(X\) is less than or equal to a specific value and also within a range.To calculate \(P(X \leq a)\), we simply evaluate the CDF at \(a\). For instance, \(P(X \leq 1)\) is calculated using \(F(1)\), which incorporates the CDF formula given for its range. This gives us a direct probability.For range probabilities such as \(P(a \leq X \leq b)\), the computation is a matter of subtracting the CDF values: \(F(b) - F(a)\). This gives the area under the PDF curve between \(a\) and \(b\), representing that probability. For example, we calculated \(P(1 \leq X \leq 3)\) using \(F(3) - F(1)\).These methods of using the CDF and PDF for probability calculations are indispensable tools in understanding and making predictions about continuous random variables.

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Most popular questions from this chapter

When circuit boards used in the manufacture of compact disc players are tested, the long-run percentage of defectives is \(5 \%\). Suppose that a batch of 250 boards has been received and that the condition of any particular board is independent of that of any other board. a. What is the approximate probability that at least \(10 \%\) of the boards in the batch are defective? b. What is the approximate probability that there are exactly ten defectives in the batch?

Let \(R\) have mean 10 and standard deviation 1.5. Find the approximate mean and standard deviation for the area of the circle with radius \(R\).

Let \(X\) denote the distance \((\mathrm{m})\) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that for bannertailed kangaroo rats, \(X\) has an exponential distribution with parameter \(\lambda=.01386\) (as suggested in the article "Competition and Dispersal from Multiple Nests,"Ecology, 1997: 873–883). a. What is the probability that the distance is at most \(100 \mathrm{~m}\) ? At most \(200 \mathrm{~m}\) ? Between 100 and \(200 \mathrm{~m}\) ? b. What is the probability that distance exceeds the mean distance by more than 2 standard deviations? c. What is the value of the median distance?

Suppose Appendix Table A.3 contained \(\Phi(z)\) only for \(z \geq 0\). Explain how you could still compute a. \(P(-1.72 \leq Z \leq-.55)\) b. \(P(-1.72 \leq Z \leq .55)\) Is it necessary to table \(\Phi(z)\) for \(z\) negative? What property of the standard normal curve justifies your answer?

Let \(X\) denote the temperature at which a certain chemical reaction takes place. Suppose that \(X\) has pdf $$ f(x)=\left\\{\begin{array}{cc} \frac{1}{9}\left(4-x^{2}\right) & -1 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Determine the cdf and sketch it. c. Is 0 the median temperature at which the reaction takes place? If not, is the median temperature smaller or larger than 0 ? d. Suppose this reaction is independently carried out once in each of ten different labs and that the pdf of reaction time in each lab is as given. Let \(Y=\) the number among the ten labs at which the temperature exceeds 1 . What kind of distribution does \(Y\) have? (Give the name and values of any parameters.)

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