What is the probability that total waiting time is at most 4 min? What is the probability that total waiting time is between 4 and 8 min? What is the probability that total waiting time is either less than 3 min
What is the probability that total waiting time is at most 4 min? What is the probability that total waiting time is between 4 and 8 min? What is the probability that total waiting time is either less than 3 min
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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What is the
What is the probability that total waiting time is between 4 and 8 min?
What is the probability that total waiting time is either less than 3 min or more than 9 min?
![In commuting to work, a professor must first get on a bus near her house and then transfer to a second bus. If the waiting time (in minutes) at each stop has a uniform distribution with \( A = 0 \) and \( B = 5 \), then it can be shown that the total waiting time \( Y \) has the probability density function (pdf) below.
\[
f(y) =
\begin{cases}
\frac{1}{25}y & 0 \leq y < 5 \\
\frac{2}{5} - \frac{1}{25}y & 5 \leq y \leq 10 \\
0 & y < 0 \text{ or } y > 10
\end{cases}
\]
**Explanation:**
The piecewise function defines the probability density function of the total waiting time \( Y \):
- For \( 0 \leq y < 5 \), the pdf is given by \( \frac{1}{25}y \), which indicates a linear increase in probability density.
- For \( 5 \leq y \leq 10 \), the pdf is \( \frac{2}{5} - \frac{1}{25}y \), showing a linear decrease in probability density.
- For \( y < 0 \) or \( y > 10 \), the pdf is 0, indicating that these waiting times are not possible.
The diagram can typically be interpreted as a triangular shape, starting from zero, rising to a peak, and then descending back to zero across the interval \( [0, 10] \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa745599b-f7be-467d-aee3-7d285fbabe65%2Fb421d2d6-dc44-4f76-8a89-b3d5a2a61b14%2F6trf88e_processed.png&w=3840&q=75)
Transcribed Image Text:In commuting to work, a professor must first get on a bus near her house and then transfer to a second bus. If the waiting time (in minutes) at each stop has a uniform distribution with \( A = 0 \) and \( B = 5 \), then it can be shown that the total waiting time \( Y \) has the probability density function (pdf) below.
\[
f(y) =
\begin{cases}
\frac{1}{25}y & 0 \leq y < 5 \\
\frac{2}{5} - \frac{1}{25}y & 5 \leq y \leq 10 \\
0 & y < 0 \text{ or } y > 10
\end{cases}
\]
**Explanation:**
The piecewise function defines the probability density function of the total waiting time \( Y \):
- For \( 0 \leq y < 5 \), the pdf is given by \( \frac{1}{25}y \), which indicates a linear increase in probability density.
- For \( 5 \leq y \leq 10 \), the pdf is \( \frac{2}{5} - \frac{1}{25}y \), showing a linear decrease in probability density.
- For \( y < 0 \) or \( y > 10 \), the pdf is 0, indicating that these waiting times are not possible.
The diagram can typically be interpreted as a triangular shape, starting from zero, rising to a peak, and then descending back to zero across the interval \( [0, 10] \).
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