
MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Question 9
Please use simple probability rules
![### Example Problem
**Problem:**
Pizza Shack usually seats 4.2 customers every 10 minutes during the lunch service. What is the likelihood it takes between 2 and 4 minutes for the next customer to arrive?
**Explanation:**
In this problem, we need to figure out the probability that the time until the next customer arrives falls between 2 and 4 minutes. Given the rate at which customers arrive (4.2 per 10 minutes), this kind of problem can be approached using the concepts of Poisson processes and exponential distributions.
1. **Average Rate of Customer Arrivals (λ):**
- The average rate λ can be calculated as 4.2 customers per 10 minutes or 0.42 customers per minute.
2. **Exponential Distribution Function:**
- For a continuous variable, the probability density function (PDF) for the exponential distribution is given by:
\[ f(t) = λe^{-λt} \]
where \( t \) is the time in minutes.
3. **Calculate the Probability:**
- To determine the probability that the time \( t \) is between 2 and 4 minutes, we need to compute the cumulative distribution function (CDF) for these time points and find the difference:
\[ P(2 \leq t \leq 4) = F(4) - F(2) \]
where \( F(t) = 1 - e^{-λt} \).
4. **Applying the Values:**
- Compute \( F(4) \) and \( F(2) \):
\[ F(4) = 1 - e^{-0.42 \times 4} = 1 - e^{-1.68} \]
\[ F(2) = 1 - e^{-0.42 \times 2} = 1 - e^{-0.84} \]
- Now find the difference:
\[ P(2 \leq t \leq 4) = (1 - e^{-1.68}) - (1 - e^{-0.84}) \]
\[ P(2 \leq t \leq 4) = e^{-0.84} - e^{-1.68} \]
By solving the above expression using a calculator, you would get the exact numerical likelihood value.
**Note:** The example](https://content.bartleby.com/qna-images/question/06a9a237-982c-465e-8a17-4be5fc663e3e/bae60c06-a686-48f9-aaed-bd7a94d0cbe0/3fxjhso_thumbnail.jpeg)
Transcribed Image Text:### Example Problem
**Problem:**
Pizza Shack usually seats 4.2 customers every 10 minutes during the lunch service. What is the likelihood it takes between 2 and 4 minutes for the next customer to arrive?
**Explanation:**
In this problem, we need to figure out the probability that the time until the next customer arrives falls between 2 and 4 minutes. Given the rate at which customers arrive (4.2 per 10 minutes), this kind of problem can be approached using the concepts of Poisson processes and exponential distributions.
1. **Average Rate of Customer Arrivals (λ):**
- The average rate λ can be calculated as 4.2 customers per 10 minutes or 0.42 customers per minute.
2. **Exponential Distribution Function:**
- For a continuous variable, the probability density function (PDF) for the exponential distribution is given by:
\[ f(t) = λe^{-λt} \]
where \( t \) is the time in minutes.
3. **Calculate the Probability:**
- To determine the probability that the time \( t \) is between 2 and 4 minutes, we need to compute the cumulative distribution function (CDF) for these time points and find the difference:
\[ P(2 \leq t \leq 4) = F(4) - F(2) \]
where \( F(t) = 1 - e^{-λt} \).
4. **Applying the Values:**
- Compute \( F(4) \) and \( F(2) \):
\[ F(4) = 1 - e^{-0.42 \times 4} = 1 - e^{-1.68} \]
\[ F(2) = 1 - e^{-0.42 \times 2} = 1 - e^{-0.84} \]
- Now find the difference:
\[ P(2 \leq t \leq 4) = (1 - e^{-1.68}) - (1 - e^{-0.84}) \]
\[ P(2 \leq t \leq 4) = e^{-0.84} - e^{-1.68} \]
By solving the above expression using a calculator, you would get the exact numerical likelihood value.
**Note:** The example
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