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The average time it takes a salesperson to finish a sale on the phone is 9 minutes and is exponentially distributed. Find the probability that more than 2 minutes pass before a sale is completed.

The average time it takes a salesperson to finish a sale on the phone is 9 minutes and is exponentially distributed. Find the probability that more than 2 minutes pass before a sale is completed.

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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2c need help pls
**Problem Statement:** The average time it takes a salesperson to finish a sale on the phone is 9 minutes and is exponentially distributed. Find the probability that more than 2 minutes pass before a sale is completed. **Solution:** Given: - Mean time (\(\lambda^{-1}\)) = 9 minutes To find the probability that more than 2 minutes pass (\(X > 2\)): The exponential probability density function is given by: \[ f(x|\lambda) = \lambda e^{-\lambda x} \] The rate parameter \(\lambda\) is the reciprocal of the mean: \[ \lambda = \frac{1}{9} \] The probability that more than 2 minutes pass is given by the survival function: \[ P(X > 2) = 1 - P(X \leq 2) = 1 - (1 - e^{-\lambda \cdot 2}) \] \[ P(X > 2) = e^{-\lambda \cdot 2} \] \[ P(X > 2) = e^{-\frac{1}{9} \cdot 2} \] \[ P(X > 2) \approx e^{-0.2222} \] Calculating: \[ P(X > 2) \approx e^{-0.2222} \approx 0.8007 \] Therefore, the probability that more than 2 minutes pass before a sale is completed is approximately 0.8007 or 80.07%.
Transcribed Image Text:**Problem Statement:** The average time it takes a salesperson to finish a sale on the phone is 9 minutes and is exponentially distributed. Find the probability that more than 2 minutes pass before a sale is completed. **Solution:** Given: - Mean time (\(\lambda^{-1}\)) = 9 minutes To find the probability that more than 2 minutes pass (\(X > 2\)): The exponential probability density function is given by: \[ f(x|\lambda) = \lambda e^{-\lambda x} \] The rate parameter \(\lambda\) is the reciprocal of the mean: \[ \lambda = \frac{1}{9} \] The probability that more than 2 minutes pass is given by the survival function: \[ P(X > 2) = 1 - P(X \leq 2) = 1 - (1 - e^{-\lambda \cdot 2}) \] \[ P(X > 2) = e^{-\lambda \cdot 2} \] \[ P(X > 2) = e^{-\frac{1}{9} \cdot 2} \] \[ P(X > 2) \approx e^{-0.2222} \] Calculating: \[ P(X > 2) \approx e^{-0.2222} \approx 0.8007 \] Therefore, the probability that more than 2 minutes pass before a sale is completed is approximately 0.8007 or 80.07%.
Expert Solution
Step 1: Introduce the given information

2c) According to the given information,

Mean, μ = 9 minutes.

Let X be a random variable represents the time pass before a sale is completed.

Using formula,

P left parenthesis X greater than x right parenthesis equals e to the power of fraction numerator negative x over denominator mu end fraction end exponent

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