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Let x have an exponential distribution with λ = 1. Find the probability. (Round your answer to four decimal places. P(x > 2)

Let x have an exponential distribution with λ = 1. Find the probability. (Round your answer to four decimal places. P(x > 2)

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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**Problem Statement:** Let \( x \) have an exponential distribution with \( \lambda = 1 \). Find the probability. (Round your answer to four decimal places.) \[ P(1 < x < 6) \] **Answer Box:** [Input box for response]
Transcribed Image Text:**Problem Statement:** Let \( x \) have an exponential distribution with \( \lambda = 1 \). Find the probability. (Round your answer to four decimal places.) \[ P(1 < x < 6) \] **Answer Box:** [Input box for response]
**Problem:** Let \( x \) have an exponential distribution with \( \lambda = 1 \). Find the probability. (Round your answer to four decimal places.) \[ P(x > 2) \] --- **Solution:** To find the probability \( P(x > 2) \), use the formula for the cumulative distribution function (CDF) of an exponential distribution: \[ P(x > a) = 1 - F(a) = 1 - (1 - e^{-\lambda a}) = e^{-\lambda a} \] For this problem, \( \lambda = 1 \) and \( a = 2 \): \[ P(x > 2) = e^{-1 \times 2} = e^{-2} \] Calculating \( e^{-2} \), we get: \[ P(x > 2) \approx 0.1353 \] Therefore, the probability that \( x \) is greater than 2 is approximately **0.1353**.
Transcribed Image Text:**Problem:** Let \( x \) have an exponential distribution with \( \lambda = 1 \). Find the probability. (Round your answer to four decimal places.) \[ P(x > 2) \] --- **Solution:** To find the probability \( P(x > 2) \), use the formula for the cumulative distribution function (CDF) of an exponential distribution: \[ P(x > a) = 1 - F(a) = 1 - (1 - e^{-\lambda a}) = e^{-\lambda a} \] For this problem, \( \lambda = 1 \) and \( a = 2 \): \[ P(x > 2) = e^{-1 \times 2} = e^{-2} \] Calculating \( e^{-2} \), we get: \[ P(x > 2) \approx 0.1353 \] Therefore, the probability that \( x \) is greater than 2 is approximately **0.1353**.
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