a passenger arriving at the Despin Spaceport must wait for a hotel shuttle is unformy between 3 and 15 minutes. (a) Sketch the density curve for the distribution of wait times. Be sure to label appropriately. (b) What is the probability a passenger waits between 4 and 12 minutes for the shuttle?
a passenger arriving at the Despin Spaceport must wait for a hotel shuttle is unformy between 3 and 15 minutes. (a) Sketch the density curve for the distribution of wait times. Be sure to label appropriately. (b) What is the probability a passenger waits between 4 and 12 minutes for the shuttle?
MATLAB: An Introduction with Applications
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ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
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Question 8 - Continuous Random Variables
Please solve the problem with simple probability rules
![### Uniform Distribution of Shuttle Wait Times
**Problem Statement:**
The amount of time a passenger arriving at the Bespin Spaceport must wait for a hotel shuttle is uniformly distributed between 3 and 15 minutes.
1. **Density Curve Sketch:**
- Sketch the density curve for the distribution of wait times. Be sure to label appropriately.
To graph the density curve for a uniformly distributed variable over the interval [3, 15]:
- On the x-axis, label the values from 3 to 15, representing the range of wait times.
- On the y-axis, the height of the uniform distribution's density function is calculated as \( \frac{1}{15-3} = \frac{1}{12} \approx 0.0833 \), as this value normalizes the area under the curve to one.
- Draw a horizontal line at \( y = 0.0833 \) extending from x = 3 to x = 15.
- Label the x-axis with "Wait Time (minutes)" and the y-axis with "Density".
2. **Probability Calculation:**
- What is the probability a passenger waits between 4 and 12 minutes for the shuttle?
For a uniform distribution, the probability \( P(a \leq X \leq b) \) is given by the proportion of the interval [a, b] over the entire range [3, 15].
Let \( a = 4 \) and \( b = 12 \):
\[
P(4 \leq X \leq 12) = \frac{(12 - 4)}{(15 - 3)} = \frac{8}{12} = \frac{2}{3} \approx 0.6667
\]
Thus, the probability that a passenger waits between 4 and 12 minutes for the shuttle is approximately 0.667 or 66.7%.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F57c1a9e8-096b-4fde-a3f9-c0ea42610dcd%2F7c5156ac-0a51-41b7-b714-ac5c93c9fb71%2F0k2wgh_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Uniform Distribution of Shuttle Wait Times
**Problem Statement:**
The amount of time a passenger arriving at the Bespin Spaceport must wait for a hotel shuttle is uniformly distributed between 3 and 15 minutes.
1. **Density Curve Sketch:**
- Sketch the density curve for the distribution of wait times. Be sure to label appropriately.
To graph the density curve for a uniformly distributed variable over the interval [3, 15]:
- On the x-axis, label the values from 3 to 15, representing the range of wait times.
- On the y-axis, the height of the uniform distribution's density function is calculated as \( \frac{1}{15-3} = \frac{1}{12} \approx 0.0833 \), as this value normalizes the area under the curve to one.
- Draw a horizontal line at \( y = 0.0833 \) extending from x = 3 to x = 15.
- Label the x-axis with "Wait Time (minutes)" and the y-axis with "Density".
2. **Probability Calculation:**
- What is the probability a passenger waits between 4 and 12 minutes for the shuttle?
For a uniform distribution, the probability \( P(a \leq X \leq b) \) is given by the proportion of the interval [a, b] over the entire range [3, 15].
Let \( a = 4 \) and \( b = 12 \):
\[
P(4 \leq X \leq 12) = \frac{(12 - 4)}{(15 - 3)} = \frac{8}{12} = \frac{2}{3} \approx 0.6667
\]
Thus, the probability that a passenger waits between 4 and 12 minutes for the shuttle is approximately 0.667 or 66.7%.
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