/*! 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. \( \approx 0.59657 \); b. \( \approx 0.36919 \); c. \( f(x) = \frac{1}{4}\ln \left(\frac{4}{x}\right) \) for \(0 < x \leq 4\).

Step by step solution

01

Write the CDF for Different Regions

The given cumulative distribution function (CDF) is: \[ F(x)=\begin{cases} 0, & x \leq 0,\ \frac{x}{4}\left(1+\ln \left(\frac{4}{x}\right)\right), & 04 \end{cases} \] This will guide our evaluation of probabilities and help derive the probability density function (PDF).
02

Find P(X ≤ 1)

Given the CDF, for \(x = 1\): \[ P(X \leq 1) = F(1) = \frac{1}{4}\left(1 + \ln \left(\frac{4}{1}\right)\right) \] Simplifying, we calculate: \[ F(1) = \frac{1}{4}(1 + \ln(4)) = \frac{1}{4}(1 + \ln(4)) = \frac{1 + \ln(4)}{4}. \] Substitute \(\ln(4)\approx 1.38629\): \[ P(X \leq 1) = \frac{1 + 1.38629}{4} \approx 0.59657. \]
03

Find P(1 ≤ X ≤ 3)

By definition, \(P(1 \leq X \leq 3) = F(3) - F(1)\). First, calculate \(F(3)\): \[ F(3) = \frac{3}{4}\left(1 + \ln \left(\frac{4}{3}\right)\right) \] Substitute \(\ln(4/3)\approx 0.28768\): \[ F(3) = \frac{3}{4}(1 + 0.28768) = \frac{3(1 + 0.28768)}{4} \approx 0.96576. \] Now calculate \(P(1 \leq X \leq 3)\): \[ P(1 \leq X \leq 3) = 0.96576 - 0.59657 \approx 0.36919. \]
04

Derive the PDF of X

The probability density function (PDF), \(f(x)\), is the derivative of the CDF, \(F(x)\), in the interval \(0 < x \leq 4\). Differentiate the CDF: \[ f(x) = \frac{d}{dx}\left(\frac{x}{4}\left(1 + \ln \left(\frac{4}{x}\right)\right)\right). \] Using the product rule and chain rule: \[ f(x) = \frac{1}{4}\left(1 + \ln \left(\frac{4}{x}\right)\right) - \frac{1}{4}. \] Therefore, the PDF in the interval \(0 < x \leq 4\) is given by: \[ f(x) = \frac{1}{4}\left(\ln \left(\frac{4}{x}\right)\right). \]

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Key Concepts

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

Probability Distribution Function
A Probability Distribution Function (PDF) is vital in statistics for defining all the possible values of a random variable and their associated probabilities. A key feature of a PDF is that it describes the likelihood of a continuous random variable taking on a particular range of values rather than specific points. For continuous variables, the probability of any single value occurring is zero; instead, areas under the curve represent probabilities.

In the exercise, we worked with both the cumulative distribution function (CDF) and derived the PDF for variable \( X \) using differentiation. This involves applying calculus because the PDF is essentially the derivative of the CDF over a specified interval. Understanding these mathematical tools is highly beneficial because they allow us to model and make predictions about real-world phenomena like hydrologic data.

Remember, one of the main properties of a PDF is that its integral over the entire space of possible outcomes equals one. This attribute signifies that the sum of probabilities for all potential outcomes collectively represents certainty.
Continuous Random Variable
A Continuous Random Variable is one that can take on an infinite number of values within a given range. The measurements of such variables usually stem from real-world observations, such as measuring time, distance, temperature, or in this case, water flow rates in hydrology.

In the problem, we defined \( X \) as a continuous random variable with specific distribution properties outlined by the CDF function. This means that \( X \) could assume any real number within its specified domain, and the function describes how probabilities are distributed across that domain. The continuous nature of \( X \) is evident in the need to use calculus to solve for probabilities, making derivatives (like the PDF) essential tools.

It's crucial to understand that working with continuous random variables involves dealing with intervals and ranges, as opposed to discrete points. This characteristic differentiates continuous variables from discrete types, which take distinct and separate values.
Hydrologic Model
A Hydrologic Model is an essential tool in understanding and predicting water-related phenomena in fields like environmental science and engineering. It typically includes mathematical functions and statistical computations to simulate real-world water systems.

In this context, the CDF described in our exercise is part of modeling hydrologic variables. Statistical functions, such as those we solved for, help model and predict water flow and distribution rates. This specific modeling is crucial as it aids in planning, predicting floods, managing water resources, and more.

Applying theoretical statistical models in hydrology can be quite challenging due to the unpredictable nature of environmental data. However, by using mathematical techniques to outline probabilities and likely outcomes—like the probability and cumulative distribution functions—we can improve our understanding and response strategies. Such models don't just aid in forecasting but also in understanding underlying processes, making them indispensable for scientists and engineers alike.

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Most popular questions from this chapter

Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime \(X\) (in weeks) has a gamma distribution with mean 24 weeks and standard deviation 12 weeks. a. What is the probability that a transistor will last between 12 and 24 weeks? b. What is the probability that a transistor will last at most 24 weeks? Is the median of the lifetime distribution less than 24 ? Why or why not? c. What is the 99th percentile of the lifetime distribution? d. Suppose the test will actually be terminated after \(t\) weeks. What value of \(t\) is such that only \(.5 \%\) of all transistors would still be operating at termination?

In each case, determine the value of the constant \(c\) that makes the probability statement correct. a. \(\Phi(c)=.9838\) b. \(P(0 \leq Z \leq c)=.291\) c. \(P(c \leq Z)=.121\) d. \(P(-c \leq Z \leq c)=.668\) e. \(P(c \leq|Z|)=.016\)

The completion time \(X\) for a certain task has cdf \(F(x)\) given by $$ \left\\{\begin{array}{cc} 0 & x<0 \\ \frac{x^{3}}{3} & 0 \leq x<1 \\ 1-\frac{1}{2}\left(\frac{7}{3}-x\right)\left(\frac{7}{4}-\frac{3}{4} x\right) & 1 \leq x \leq \frac{7}{3} \\ 1 & x>\frac{7}{3} \end{array}\right. $$ a. Obtain the pdf \(f(x)\) and sketch its graph. b. Compute \(P(.5 \leq X \leq 2)\). c. Compute \(E(X)\).

The authors of the article "A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses" (IEEE Trans. on Elect. Insulation, 1985: 519-522) state that "the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress." They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens subjected to \(\mathrm{AC}\) voltage. The values of the parameters depend on the voltage and temperature; suppose \(\alpha=2.5\) and \(\beta=200\) (values suggested by data in the article). a. What is the probability that a specimen's lifetime is at most 250? Less than 250? More than 300 ? b. What is the probability that a specimen's lifetime is between 100 and 250 ? c. What value is such that exactly \(50 \%\) of all specimens have lifetimes exceeding that value?

Let \(X\) denote the data transfer time (ms) in a grid computing system (the time required for data transfer between a "worker" computer and a "master" computer. Suppose that \(X\) has a gamma distribution with mean value \(37.5 \mathrm{~ms}\) and standard deviation \(21.6\) (suggested by the article "Computation Time of Grid Computing with Data Transfer Times that Follow a Gamma Distribution,' Proceedings of the First International Conference on Semantics, Knowledge, and Grid, 2005). a. What are the values of \(\alpha\) and \(\beta\) ? b. What is the probability that data transfer time exceeds \(50 \mathrm{~ms}\) ? c. What is the probability that data transfer time is between 50 and \(75 \mathrm{~ms}\) ?

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