91Ó°ÊÓ

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 pdf $$ f(y)=\left\\{\begin{array}{cl} \frac{1}{25} y & 0 \leq y<5 \\ \frac{2}{5}-\frac{1}{25} y & 5 \leq y \leq 10 \\ 0 & y<0 \text { or } y>10 \end{array}\right. $$ a. Sketch a graph of the pdf of \(Y\). b. Verify that \(\int_{-\infty}^{\infty} f(y) d y=1\). c. What is the probability that total waiting time is at most 3 min? d. What is the probability that total waiting time is at most 8 min? e. What is the probability that total waiting time is between 3 and \(8 \mathrm{~min}\) ? f. What is the probability that total waiting time is either less than 2 min or more than 6 min?

Step by step solution

01

Understanding the PDF

The given probability density function, (pdf), which describes the behavior of the total waiting time, is piecewise. It includes two main sections based on the value of total waiting time \( Y \):- From 0 to 5 minutes: \( f(y) = \frac{1}{25}y \)- From 5 to 10 minutes: \( f(y) = \frac{2}{5} - \frac{1}{25}y \)Outside these ranges, \( f(y) = 0 \).

02

Sketch the PDF

To sketch this pdf, plot \( f(y) \) against \( y \):- Draw a line with a positive slope from (0,0) to (5,0.2).- Then draw a line with a negative slope from (5,0.2) to (10,0).Beyond these values, the function remains at 0.

03

Verify Total Probability

To ensure this pdf represents a valid probability distribution, integrate \( f(y) \) across its entire domain, ensuring the total area equals 1.- \[ \int_{0}^{5} \frac{1}{25}y \, dy = \frac{1}{50}y^2 \Bigg|_0^5 = 0.5 \]- \[ \int_{5}^{10} \left(\frac{2}{5} - \frac{1}{25}y\right) \, dy = \left(\frac{2}{5}y - \frac{1}{50}y^2\right) \Bigg|_5^{10} = 0.5 \]- Total integral = 0.5 + 0.5 = 1, confirming it's a valid pdf.

04

Calculate Probability for At Most 3 Minutes

The probability that \( Y \leq 3 \) is found by integrating the pdf from 0 to 3.- \[ \int_{0}^{3} \frac{1}{25}y \, dy = \frac{1}{50}y^2 \Bigg|_0^3 = \frac{1}{50}(9) = 0.18 \]

05

Calculate Probability for At Most 8 Minutes

To calculate \( P(Y \leq 8) \), integrate:- From 0 to 5: \( \int_{0}^{5} \frac{1}{25}y \, dy = 0.5 \)- From 5 to 8: \[ \int_{5}^{8} \left(\frac{2}{5} - \frac{1}{25}y\right) \, dy = \left(\frac{2}{5}y - \frac{1}{50}y^2\right) \Bigg|_5^{8} = 0.42 \]- Thus, \( P(Y \leq 8) = 0.5 + 0.42 = 0.92 \).

06

Calculate Probability for Between 3 and 8 Minutes

For \( P(3 < Y < 8) \), integrate separately and subtract:- \[ P(Y \leq 8) = 0.92 \]- \[ P(Y \leq 3) = 0.18 \]- Thus, \( P(3 < Y < 8) = 0.92 - 0.18 = 0.74 \).

07

Calculate Probability for Either Less Than 2 or More Than 6 Minutes

We break it into two integrals for probability calculation:- For \( P(Y < 2) \): - \[ \int_{0}^{2} \frac{1}{25}y \, dy = \frac{1}{50}(4) = 0.08 \]- For \( P(Y > 6) \): Combine the ranges from \( 6 \) to \( 10 \). - From 6 to 10: \[ \int_{6}^{10} \left(\frac{2}{5} - \frac{1}{25}y\right) \, dy = 0.32 \]- Add probabilities: \( P(Y < 2 \, or \, Y > 6) = 0.08 + 0.32 = 0.4 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Uniform Distribution

A uniform distribution is a type of continuous probability distribution where all outcomes are equally likely. Imagine you're rolling a fair six-sided die. Each side has an equal chance of landing face up. In mathematical terms, if the outcomes are spread evenly over the range from \(A = 0\) to \(B = 5\), it means every value between \(0\) and \(5\) has the same probability.The key features of a uniform distribution:

• Defined by two parameters: the minimum value \(A\) and the maximum value \(B\).
• Probability density function (pdf) is constant between \(A\) and \(B\).

For our example, each bus stop waiting time is modeled with uniform distribution where each moment from \(0\) to \(5\) minutes has the same probability density. This simplifies calculations because the total waiting time \(Y\) must also follow a predictable piecewise function over its total range which combines two stops.

Probability Density Function (pdf)

The probability density function (pdf) helps determine probabilities for continuous distributions by associating potential values with likelihoods. It's not probability directly, but a way to represent data that can be converted into probabilities through integration.In our problem, the pdf of the total waiting time \(Y\) is given piecewise:

• From \(0\) to \(5\) minutes, \( f(y) = \frac{1}{25}y \).
• From \(5\) to \(10\) minutes, \( f(y) = \frac{2}{5} - \frac{1}{25}y \).
• Outside these ranges, \(f(y) = 0\).

This piecewise function represents how the probabilities change over different intervals of \(Y\). Graphically, if we sketch this function, we begin with a linearly increasing section up to \(5\) minutes, peaking there, then linearly decreasing to \(10\) minutes. This visual aids in understanding where \(Y\) is most likely to occur, peaking at \(5\) minutes.

Integration of pdf

Integration of the pdf is a fundamental process for finding probabilities in continuous distributions. We calculate the area under the pdf curve to determine the probability that the variable \(Y\) falls within a specific range.Here's how we approach integration for our pdf:

• Verify that integration over the entire domain yields a total probability of 1, confirming a valid pdf. This involves adding \[\int_{0}^{5} \frac{1}{25}y \, dy + \int_{5}^{10} \left(\frac{2}{5} - \frac{1}{25}y\right) \, dy = 1\]
• To find the probability of intervals, like \(Y \leq 8\), we calculate:\[\int_{0}^{5} \frac{1}{25}y \, dy + \int_{5}^{8} \left(\frac{2}{5} - \frac{1}{25}y\right) \, dy\]Resulting in the cumulative probability over that range.

Integration not only solidifies the understanding of the pdf but connects the dots between the abstract pdf and tangible probabilities, such as the likelihood of a commuter waiting less than 8 minutes.