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The article 'Error Distribution in Navigation"' \((J\). of the Institute of Navigation, 1971: 429-442) suggests that the frequency distribution of positive errors (magnitudes of errors) is well approximated by an exponential distribution. Let \(X=\) the lateral position error (nautical miles), which can be either negative or positive. Suppose the pdf of \(X\) is $$ f(x)=(.1) e^{-.2|x|}-\infty

Short Answer

Expert verified
PDF is legitimate; CDF and probabilities calculated. PDF symmetric around zero; CDF starts at 0 and rises to 1.

Step by step solution

01

Understand the PDF

The given PDF is \( f(x) = 0.1 e^{-0.2|x|} \) for \(-\infty < x < \infty\). This is an exponential distribution modified to accommodate both positive and negative values of \(x\) by taking the absolute value.
02

Check PDF Legitimacy

To verify that \( f(x) \) is a legitimate PDF, we need to show that it integrates to 1 over its range. Since \( |x| \) splits the function symmetrically, calculate the integral of \( 0.1 e^{-0.2|x|} \) from \(-\infty\) to \(\infty\). This can be simplified into two integrals, from \(-\infty\) to 0 and from 0 to \(\infty\):\[\int_{-\infty}^{0} 0.1 e^{0.2x} \, dx + \int_{0}^{\infty} 0.1 e^{-0.2x} \, dx= \frac{0.1}{0.2} + \frac{0.1}{0.2} = 1.\]The integral indeed sums up to 1, confirming \( f(x) \) is a valid PDF.
03

Sketch the PDF

The graph of \( f(x) = 0.1 e^{-0.2|x|} \) is symmetrically shaped like two exponential decays with a peak at \( x = 0 \) (the origin), decreasing towards both \( -\infty \) and \( \infty \).
04

Derive the CDF

To get the CDF, integrate the PDF from \(-\infty\) to \(x\): For \( x \geq 0: P(X \leq x) = \int_{-\infty}^{x} 0.1 e^{-0.2|t|} \, dt = 0.5 + 0.5(1-e^{-0.2x}).\)For \( x < 0: P(X \leq x) = \int_{-\infty}^{x} 0.1 e^{-0.2|t|} \, dt = \frac{1}{2}e^{0.2x}.\)
05

Sketch the CDF

The CDF graph starts at 0 for \( x=-\infty \), monotonically increases to 1 and is symmetric about the origin; at \( x=0 \), the value is 0.5, increases gradually as \(x \to \infty\).
06

Calculate Probabilities

Calculate each probability using the CDF:- **\( P(X \leq 0) \)**: This is simply the CDF at 0, which is 0.5.- **\( P(X \leq 2) \)**: Use the CDF formula for \( X \geq 0 \), \( P(X \leq 2) = 0.5 + 0.5(1-e^{-0.4}) \approx 0.6636. \)- **\( P(-1 \leq X \leq 2) \)**: \( P(X \leq 2) - P(X \leq -1) = 0.6636 - 0.5e^{0.2} \approx 0.4622 \)- **Error more than 2 miles**: Find \( 1 - P(X \leq 2) = 1 - 0.6636 = 0.3364 \).

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Key Concepts

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

Probability Density Function (PDF)
The probability density function (PDF) is a fundamental concept in statistics used to specify the probability of a random variable falling within a particular range of values. For continuous random variables, the PDF provides the height of the curve at any given point along the distribution. This specific curve does not give probabilities directly, but rather represents the entire probability distribution.

In the context of the exponential distribution, the PDF is particularly useful as it describes the likelihood of random events and can model time until an event occurs. For this exercise, the PDF is given by:
  • \( f(x) = 0.1 e^{-0.2|x|} \) where \(-\infty < x < \infty\).
**Key Characteristics**:
  • The PDF is non-negative for all values of \(x\).
  • The total area under the curve of the PDF is 1, confirming it represents a legitimate probability distribution.
  • This specific PDF is modified by the absolute value notation \(|x|\), making it suitable to accommodate both negative and positive errors.
To check if a function is a proper PDF, you integrate it over its entire range. If the result is 1, it is confirmed to represent a valid probability distribution.
Cumulative Distribution Function (CDF)
The cumulative distribution function (CDF) is another fundamental tool in probability and statistics. It calculates the probability that a random variable takes on a value less than or equal to a certain amount. For continuous random variables, the CDF is derived by integrating the PDF from the start of the distribution up to point \(x\).

In our exercise, we are dealing with a distribution where the CDF can be expressed as:
  • For \(x \geq 0:\) \(P(X \leq x) = 0.5 + 0.5(1-e^{-0.2x})\)
  • For \(x < 0:\) \(P(X \leq x) = \frac{1}{2}e^{0.2x}\)
**Properties of the CDF include**:
  • Starts from 0 and increases to 1 as \(x\) moves from \(-\infty\) to \(+\infty\).
  • It is a non-decreasing function, which means once it reaches a certain probability, it cannot go down.
  • The graph is smooth and reflects the symmetric nature of the errors about the origin, with a value of 0.5 at x=0.
To visualize the CDF, imagine a curve starting from 0 on the left and smoothly rising to 1 towards the right. At \(x=0\), the CDF is exactly 0.5, and it gradually approaches 1 as \(x\) approaches positive infinity.
Probability Calculations
Calculating probabilities using the PDF and CDF allows us to understand the likelihood of different scenarios occurring within the context of the distribution. In the context of this specific exercise, different probability queries are solved using the CDF.

Let's break down some common calculations involved:
  • **\(P(X \leq 0)\):** This represents the probability that the random variable \(X\) is zero or negative. It is equal to the CDF at zero, which is 0.5.
  • **\(P(X \leq 2)\):** To find this probability, use the CDF formula for non-negative X. The calculation results in a probability of roughly 0.6636.
  • **\(P(-1 \leq X \leq 2)\):** This involves calculating the difference of probabilities. Use the CDF values: \(P(X \leq 2)\) minus \(P(X \leq -1)\), resulting in approximately 0.4622.
  • **Probability of an Error Greater Than 2 miles:** Use the complement rule. The calculation \(1 - P(X \leq 2)\) gives a probability of about 0.3364.
The above procedures highlight how the CDF aids in determining probabilities over specified intervals, leveraging the properties of the exponential distribution.

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Most popular questions from this chapter

Let \(X\) represent the number of individuals who respond to a particular online coupon offer. Suppose that \(X\) has approximately a Weibull distribution with \(\alpha=10\) and \(\beta=20\). Calculate the best possible approximation to the probability that \(X\) is between 15 and 20 , inclusive.

The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value \(30 \mathrm{~mm}\) and standard deviation \(7.8 \mathrm{~mm}\) [suggested in the article "Reliability Evaluation of Corroding Pipelines Considering Multiple Failure Modes and TimeDependent Internal Pressure" \((J\). of Infrastructure Systems, 2011: 216-224)]. a. What is the probability that defect length is at most \(20 \mathrm{~mm}\) ? Less than \(20 \mathrm{~mm}\) ? b. What is the 75 th percentile of the defect length distribution-that is, the value that separates the smallest \(75 \%\) of all lengths from the largest \(25 \%\) ? c. What is the 15 th percentile of the defect length distribution? d. What values separate the middle \(80 \%\) of the defect length distribution from the smallest \(10 \%\) and the largest \(10 \%\) ?

The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cl} 2\left(1-\frac{1}{x^{2}}\right) & 1 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Compute the cdf of \(X\). b. Obtain an expression for the \((100 p)\) th percentile. What is the value of \(\tilde{\mu}\) ? c. Compute \(E(X)\) and \(V(X)\). d. If \(1.5\) thousand gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the \(1.5\) thousand gallons is expected to be left at the end of the week? [Hint: Let \(h(x)=\) amount left when demand \(=x\).]

Let \(X\) have a uniform distribution on the interval \([A, B]\). a. Obtain an expression for the \((100 p)\) th percentile. b. Compute \(E(X), V(X)\), and \(\sigma_{X}\). c. For \(n\), a positive integer, compute \(E\left(X^{n}\right)\).

The article "Three Sisters Give Birth on the Same Day" (Chance, Spring 2001, 23-25) used the fact that three Utah sisters had all given birth on March 11, 1998 as a basis for posing some interesting questions regarding birth coincidences. a. Disregarding leap year and assuming that the other 365 days are equally likely, what is the probability that three randomly selected births all occur on March 11? Be sure to indicate what, if any, extra assumptions you are making. b. With the assumptions used in part (a), what is the probability that three randomly selected births all occur on the same day? c. The author suggested that, based on extensive data, the length of gestation (time between conception and birth) could be modeled as having a normal distribution with mean value 280 days and standard deviation \(19.88\) days. The due dates for the three Utah sisters were March 15, April 1, and April 4, respectively. Assuming that all three due dates are at the mean of the distribution, what is the probability that all births occurred on March 11? d. Explain how you would use the information in part (c) to calculate the probability of a common birth date.

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