/*! 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 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 certain 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. 0.5965 b. 0.3695 c. PDF is \( \frac{1}{4}\left[\ln\left(\frac{4}{x}\right) - 1\right] \) for \(0 < x \leq 4\)

Step by step solution

01

Analyze the Given CDF

The cumulative distribution function (CDF) for a continuous random variable (rv) is given piecewise, defining three regions: 1. For \( x \leq 0 \), \( F(x) = 0 \).2. For \( 0 < x \leq 4 \), \( F(x) = \frac{x}{4}\left[1+\ln\left(\frac{4}{x}\right)\right] \).3. For \( x > 4 \), \( F(x) = 1 \).To solve each question, identify which region the calculations refer to based on the value of \( x \).
02

Calculate P(X ≤ 1) for Part a

For \( P(X \leq 1) \), use the CDF formula for \( 0 < x \leq 4 \):\[F(1) = \frac{1}{4}\left[1+\ln\left(\frac{4}{1}\right)\right] = \frac{1}{4}\left[1+\ln(4)\right].\]Compute \( \ln(4) = \ln(2^2) = 2\ln(2) \approx 2 \times 0.693 = 1.386 \), so:\[F(1) = \frac{1}{4}(1 + 1.386) = \frac{2.386}{4} = 0.5965.\]
03

Calculate P(1 ≤ X ≤ 3) for Part b

This probability is the difference between the CDF values at \( x = 3 \) and \( x = 1 \): \[P(1 \leq X \leq 3) = F(3) - F(1).\]First, compute \( F(3) \) using the CDF formula for \( 0 < x \leq 4 \):\[F(3) = \frac{3}{4}\left[1+\ln\left(\frac{4}{3}\right)\right].\]Calculate \( \ln\left(\frac{4}{3}\right) \approx 0.288 \), so:\[F(3) = \frac{3}{4}(1 + 0.288) = \frac{3.864}{4} = 0.966.\]Then the probability is:\[P(1 \leq X \leq 3) = 0.966 - 0.5965 = 0.3695.\]
04

Determine the PDF from the CDF for Part c

The probability density function (PDF) is the derivative of the CDF. For \( 0 < x \leq 4 \), calculate:\[\text{If } F(x) = \frac{x}{4}\left[1 + \ln\left(\frac{4}{x}\right)\right],\]then the derivative:\(\frac{d}{dx}\left[\frac{x}{4}\left(1 + \ln\left(\frac{4}{x}\right)\right)\right] = \frac{1}{4}\left(1 + \ln\left(\frac{4}{x}\right)\right) - \frac{1}{4} \frac{x}{x} = \frac{1}{4}\ln\left(\frac{4}{x}\right).\) Thus, the PDF, \( f(x) \), for \( 0 < x \leq 4 \) is: \[f(x) = \frac{1}{4}\left[\ln\left(\frac{4}{x}\right) - 1\right].\]

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Key Concepts

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

Probability Density Function (PDF)
A Probability Density Function (PDF) is a fundamental concept in probability theory and statistics. It characterizes the likelihood of a continuous random variable taking on a specific value. Unlike discrete variables, continuous variables can take on any value in a given range, which requires a different way of representing probabilities. The PDF allows for this by providing a formula that, when integrated over a particular interval, gives the probability that the variable falls within that range.
The PDF is derived from the Cumulative Distribution Function (CDF) by differentiating it. Here, for the variable \(X\) within the interval \(0 < x \leq 4\), the CDF is given by:\[F(x) = \frac{x}{4}\left[1 + \ln\left(\frac{4}{x}\right)\right]\].
By taking its derivative, we find the PDF, which is:\[f(x) = \frac{1}{4}\left[\ln\left(\frac{4}{x}\right) - 1\right]\] for the same range. This formula tells us how the probability is distributed over different values within this interval. Note that the area under the PDF curve over any interval will give you the probability of the random variable falling within that interval.
Continuous Random Variable
A Continuous Random Variable differs from a discrete variable in that it can assume an infinite number of values within a given range. In real-world scenarios, many quantities such as height, temperature, or time can take on continuous values. This allows for more precise modeling of real-life phenomena.
In our present problem involving the variable \(X\), the CDF outlines how \(X\) behaves over different intervals. For \(x \leq 0\), \(F(x) = 0\) signifies that there's zero probability of \(X\) assuming a value less than or equal to zero. As we move to the interval \(0 < x \leq 4\), the CDF defines the probability as a function of \(x\), capturing the behavior of our continuous random variable in this range.
Finally, for \(x > 4\), \(F(x) = 1\), meaning all values of \(X\) are covered under this cumulative probability, portraying the classic property of CDFs where they converge to a total probability of one.
Probability Calculation
In probability calculations involving continuous random variables, the CDF becomes immensely useful. It helps us compute probabilities over different intervals by looking at the values of the CDF at specific points. These are depicted as differences between CDF values, translating into the probabilities of the variable landing in those intervals.
For instance, to compute \(P(X \leq 1)\), the CDF value is used directly since it's the probability \(X\) is less than or equal to 1. Here, using the derived CDF formula, you calculate: \[F(1) = \frac{1}{4}\left[1+\ln(4)\right] = 0.5965\], establishing the likelihood of \(X\) being less than 1.
Similarly, for \(P(1 \leq X \leq 3)\), you take the difference \(F(3) - F(1)\), which gives you \(0.966 - 0.5965 = 0.3695\). This highlights the probability of \(X\) being between 1 and 3. These calculations exemplify how probabilities for continuous distributions are summed up iteratively over the range, allowing us to capture the variable's stochastic behavior accurately.

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Most popular questions from this chapter

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