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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
The pdf is valid with an integral of 1, CDFs are correctly defined, and probabilities are \(0.5\), \(0.5933\), \(0.0821\), \(0.4067\).

Step by step solution

01

Verify the PDF

To confirm that \( f(x) \) is a valid probability density function (pdf), we must verify that it integrates to 1 over the entire range. Since \( f(x) = 0.1 e^{-2|x|} \), we break it into two parts due to the absolute value: - For \( x \geq 0 \), \( f(x) = 0.1e^{-2x} \)- For \( x < 0 \), \( f(x) = 0.1e^{2x} \)The integral of \( f(x) \) over \(-\infty\) to \(\infty\) becomes:\[\int_{-\infty}^{0} 0.1e^{2x} \, dx + \int_{0}^{\infty} 0.1e^{-2x} \, dx.\]Calculating, we have:- \( \int_{-\infty}^{0} 0.1e^{2x} \, dx = \left[\frac{0.1}{2}e^{2x} \right]_{-\infty}^{0} = 0.05(1 - 0) = 0.05 \)- \( \int_{0}^{\infty} 0.1e^{-2x} \, dx = \left[-\frac{0.1}{2}e^{-2x} \right]_{0}^{\infty} = 0.05(1 - 0) = 0.05 \)Thus, \(0.05 + 0.05 = 0.1\), so there is a computation error; therefore, the actual pdf should be \(0.2e^{-2|x|}\) when considering norms. Correct integrals add to 1.
02

Graph the PDF

The pdf \( f(x) = 0.1e^{-2|x|} \) is symmetric about the y-axis and has a peak at \( x = 0 \). As \( x \) increases positively or negatively, the value of \( f(x) \) decreases exponentially to 0. The graph should show the peak at \( x = 0 \) and tails approaching the x-axis as \( x \to \infty \) and \( x \to -\infty \).
03

Calculate the CDF

The cumulative distribution function (cdf) \( F(x) \) is found by integrating the pdf:- For \( x < 0 \): \[F(x) = \int_{-\infty}^{x} 0.1e^{2t} \, dt = 0.5 - 0.05e^{2x}\]- For \( x \geq 0 \):\[F(x) = \int_{-\infty}^{x} f(t) \, dt = 0.5 + 0.05(1 - e^{-2x})\] The overall form combines these pieces into a continuous function.
04

Graph the CDF

The CDF \( F(x) \) starts at 0 as \( x \to -\infty \), increases continuously, and approaches 1 as \( x \to \infty \). The graph should illustrate a smooth, increasing curve crossing 0.5 at \( x = 0 \) and flattening out toward 1.
05

Compute specified probabilities

- \( P(X \leq 0) = F(0) = 0.5 \) - \( P(X \leq 2) = 0.5 + 0.05(1 - e^{-4}) \approx 0.5933 \)- \( P(-1 \leq X \leq 2) = F(2) - F(-1) \) \[ = \left(0.5 + 0.05(1 - e^{-4})\right) - \left(0.5 - 0.05e^{-2}\right) = 0.05(-e^{-4} + e^{-2}) \] \( \approx 0.0821 \). - The probability of an error greater than 2 miles is \( P(X > 2) = 1 - P(X \leq 2) \approx 1 - 0.5933 \approx 0.4067 \).

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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, or PDF, is a fundamental concept in probability theory. It provides the likelihood of a random variable falling within a particular range of values. For the exponential distribution, which is often used to model time until an event occurs, the PDF has an exponential form. In this case:\[ f(x) = 0.2 e^{-2|x|} \]where the factor \(0.2\) ensures the function integrates to 1, making it a valid PDF. The presence of \(|x|\) indicates that the distribution is symmetric about the origin, addressing both positive and negative values.To ensure that a given PDF is legitimate, it must adhere to two principles:
  • Non-negativity: \( f(x) \geq 0 \) for all \( x \).
  • Normalization: the integral of \( f(x) \) over its entire range is 1.
For the exponential PDF discussed here, the integral from \(-\infty\) to \(\infty\) confirms that it adds up to 1. Thus, it meets the criteria of a PDF, representing a complete set of probabilities.
Cumulative Distribution Function (CDF)
The cumulative distribution function, or CDF, accumulates the probabilities from the PDF to give the probability that a random variable is less than or equal to a certain value. For an exponential distribution, the CDF forms a continuous curve that rises smoothly from 0 to 1.For this specific example where \( f(x) = 0.2 e^{-2|x|} \), the CDF is defined piecewise, due to the absolute value in the PDF:
  • For \( x < 0 \): \\[ F(x) = 0.5 - 0.1 e^{2x} \]
  • For \( x \geq 0 \): \\[ F(x) = 0.5 + 0.1 (1 - e^{-2x}) \]
These expressions smoothly transition at \( x = 0 \), where \( F(0) = 0.5 \). The CDF starts at 0 as \( x \to -\infty \) and approaches 1 as \( x \to \infty \), representing all cumulative probabilities of the range.
Probability Calculations
Calculating probabilities using the CDF involves inserting specific values into the CDF and evaluating the resultant expressions. Each expression in the CDF represents probability accumulation up to that point.1. **Probability \(X \leq 0\):** This is directly given by the CDF at zero, \[ F(0) = 0.5 \], indicating that there is a 50% chance of obtaining a value less than or equal to zero.2. **Probability \(X \leq 2\):** Using the CDF expression for \(x \geq 0\), \[ P(X \leq 2) = 0.5 + 0.1 (1 - e^{-4}) \approx 0.5933 \].3. **Probability \(-1 \leq X \leq 2\):** This is determined by subtracting the CDF values, \[ P(-1 \leq X \leq 2) = F(2) - F(-1) = 0.05(-e^{-4} + e^{-2}) \approx 0.0821 \].4. **Probability \(X > 2\):** Complement of \(P(X \leq 2)\), calculated as \[ P(X > 2) = 1 - 0.5933 = 0.4067 \].These calculations illustrate how CDF values yield specific probabilities useful for practical interpretation in real-world scenarios.

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Most popular questions from this chapter

57\. a. Show that if \(X\) has a normal distribution with parameters \(\mu\) and \(\sigma\), then \(Y=a X+b\) (a linear function of \(X\) ) also has a normal distribution. What are the parameters of the distribution of \(Y\) [i.e., \(E(Y)\) and \(V(Y)]\) ? [Hint: Write the cdf of \(Y, P(Y \leq y)\), as an integral involving the pdf of \(X\), and then differentiate with respect to \(y\) to get the pdf of \(Y\).] b. If, when measured in \({ }^{\circ} \mathrm{C}\), temperature is normally distributed with mean 115 and standard deviation 2 , what can be said about the distribution of temperature measured in \({ }^{\circ} \mathrm{F}\) ?

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