/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 In commuting to work, a professo... [FREE SOLUTION] | 91Ó°ÊÓ

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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 $$ f(y)= \begin{cases}\frac{1}{25} y & 0 \leq y<5 \\ \frac{2}{5}-\frac{1}{25} y & 5 \leq y \leq 10 \\ 0 & y<0 \text { or } y>10\end{cases} $$ 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 \(2 \mathrm{~min}\) or more than \(6 \mathrm{~min}\) ?

Short Answer

Expert verified
The total waiting time pdf integrates to 1. The probability for \(Y \leq 3\) is 0.18, \(Y \leq 8\) is 0.76, \(3 < Y \leq 8\) is 0.58, and \(Y < 2\) or \(Y > 6\) is 0.44.

Step by step solution

01

Understanding the PDF

The probability density function (pdf) of the total waiting time \(Y\) is defined differently over three distinct intervals of \(Y\). For \(0 \leq y < 5\), \(f(y) = \frac{1}{25} y\), for \(5 \leq y \leq 10\), \(f(y) = \frac{2}{5}-\frac{1}{25} y\), and \(f(y) = 0\) otherwise. It's essential to understand this definition to work through the problem.
02

Sketching the Graph of f(y)

To sketch the graph, plot \(f(y) = \frac{1}{25} y\) as a straight line from \((0, 0)\) to \((5, 0.2)\), and then \(f(y) = \frac{2}{5}-\frac{1}{25} y\) as a declining line from \((5, 0.2)\) to \((10, 0)\). These lines represent the pdf from 0 to 10, and \(f(y) = 0\) outside this range, indicating no probability density there.
03

Verify the Total Area Under the Curve is 1

Calculate the integral to verify that the total area under the pdf is 1: \(\int_{0}^{5} \frac{1}{25} y \, dy + \int_{5}^{10} \left(\frac{2}{5} - \frac{1}{25} y\right) \, dy = 1\). Evaluate these integrals separately and sum them to confirm the total area equals 1.
04

Probability of Total Waiting Time at Most 3 Min

Calculate \(P(Y \leq 3)\) by integrating \(f(y)\) from 0 to 3: \(\int_{0}^{3} \frac{1}{25} y \, dy\). The result will give the probability.
05

Probability of Total Waiting Time at Most 8 Min

Calculate \(P(Y \leq 8)\) by integrating \(f(y)\) from 0 to 5 and then from 5 to 8: \(\int_{0}^{5} \frac{1}{25} y \, dy + \int_{5}^{8} \left(\frac{2}{5} - \frac{1}{25} y\right) \, dy\). Add these values to get the total probability.
06

Probability Between 3 and 8 Min

For \(P(3 < Y \leq 8)\), compute \(\int_{3}^{5} \frac{1}{25} y \, dy + \int_{5}^{8} \left(\frac{2}{5} - \frac{1}{25} y\right) \, dy\). This yields the probability of waiting time being in the range between 3 and 8 minutes.
07

Probability Less Than 2 Min or More Than 6 Min

Calculate the probability for \(Y < 2\) and \(Y > 6\). First compute \(\int_{0}^{2} \frac{1}{25} y \, dy\), then \(1 - \int_{0}^{6} f(y) \, dy\) for the complement region. Sum both probabilities.

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Key Concepts

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

Probability Density Function
When dealing with continuous random variables, the probability density function (pdf) is an essential concept in probability and statistics. The pdf describes how the probabilities are distributed over the values of the random variable. For any value \( y \), the pdf, \( f(y) \), specifies the relative likelihood of \( Y \) being close to \( y \).

In uniform distribution, the pdf is often flat (constant) over the interval, but this example shows a case with a piecewise linear pdf. For the total waiting time \( Y \), the pdf is defined over different intervals, explicitly outlining how probable each waiting time is.

- For \( 0 \leq y < 5 \), the pdf is an increasing linear function: \( f(y) = \frac{1}{25} y \). This indicates that as \( y \) approaches 5, the likelihood of waiting time \( y \) increases.
- For \( 5 \leq y \leq 10 \), the pdf is a decreasing linear function: \( f(y) = \frac{2}{5} - \frac{1}{25} y \). Here, the likelihood starts high at \( y=5 \) but declines as \( y \) approaches 10.
- Outside of the range \([0, 10]\), the pdf is zero, meaning there's no chance of waiting time being less than 0 or more than 10 minutes.

Visualizing the pdf can help in understanding and calculating probabilities related to the waiting times. Each segment of this function tells us how the waiting times are distributed across different intervals.
Integration in Probability
Integration is a critical operation in probability, especially when working with continuous probability distributions. For a probability density function, integrating over an interval gives the cumulative probability that the random variable falls within that range.

- The total probability across all possible outcomes must add up to 1. This is expressed mathematically as \( \int_{-\infty}^{\infty} f(y) \, dy = 1 \). This guarantees that the entire area under the pdf curve is equal to 1, ensuring the function is a valid probability distribution.

In practical terms, when we integrate the pdf of the total waiting time \( Y \) over certain limits, we determine the probability that \( Y \) falls within those bounds.

For this exercise:
- To verify that the pdf is valid, check if \( \int_{0}^{10} f(y) \, dy = 1 \). This involves partitioning the integration over the two different linear expressions in the pdf: from 0 to 5 and 5 to 10.

Each integration corresponds to a specific segment of the distribution, and when combined, they confirm the completeness of the distribution. Such integrations are fundamental for calculating any specific probabilities, such as those asking for time windows or specific conditions.
Cumulative Probability
Cumulative probability reflects the total probability that a random variable is less than or equal to a certain value. In mathematical terms, the cumulative distribution function (CDF) quantifies \( P(Y \leq a) \), where \( a \) is a specific waiting time.

In this context, cumulative probability is calculated by integrating the pdf up to the desired point.
Here are examples of how to derive specific cumulative probabilities for given time intervals:

- **Probability for Waiting Time \(\leq 3 \) Minutes:** Calculate \( \int_{0}^{3} \frac{1}{25} y \, dy \). This will yield the likelihood that \( Y \) is 3 minutes or less.
- **Probability for Waiting Time \(\leq 8 \) Minutes:** Compute \( \int_{0}^{5} \frac{1}{25} y \, dy + \int_{5}^{8} \left(\frac{2}{5} - \frac{1}{25} y\right) \, dy \). This tells you the chance \( Y \) is at most 8 minutes.

The integration limits dictate the span over which we are interested. By calculating these integrals, we aggregate probabilities incrementally, leading to a comprehensive understanding of potential outcomes based on the waiting time.

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Most popular questions from this chapter

The weekly demand for propane gas (in 1000s of gallons) from a particular facility is an rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cl} 2\left(1-\frac{1}{x^{2}}\right) & 1 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Compute the cdf of \(X\). b. Obtain an expression for the \((100 p)\) th percentile. What is the value of \(\tilde{\mu}\) ? c. Compute \(E(X)\) and \(V(X)\). d. If \(1.5\) thousand gallons are in stock at the beginning of the week and no new supply is due in during the week, how much of the \(1.5\) thousand gallons is expected to be left at the end of the week? [Hint: Let \(h(x)=\) amount left when demand \(=x\).]

Let \(X\) represent the number of individuals who respond to a particular online coupon offer. Suppose that \(X\) has approximately a Weibull distribution with \(\alpha=10\) and \(\beta=20\). Calculate the best possible approximation to the probability that \(X\) is between 15 and 20 , inclusive.

Let \(Z\) be a standard normal random variable and calculate the following probabilities, drawing pictures wherever appropriate. a. \(P(0 \leq Z \leq 2.17)\) b. \(P(0 \leq Z \leq 1)\) c. \(P(-2.50 \leq Z \leq 0)\) d. \(P(-2.50 \leq Z \leq 2.50)\) e. \(P(Z \leq 1.37)\) f. \(P(-1.75 \leq Z)\) g. \(P(-1.50 \leq Z \leq 2.00)\) h. \(P(1.37 \leq Z \leq 2.50)\) i. \(P(1.50 \leq Z)\) j. \(\quad P(|Z| \leq 2.50)\)

Consider babies born in the "normal" range of \(37-43\) weeks gestational age. Extensive data supports the assumption that for such babies born in the United States, birth weight is normally distributed with mean \(3432 \mathrm{~g}\) and standard deviation \(482 \mathrm{~g}\). [The article "Are Babies Normal?"' (The American Statistician, 1999: 298-302) analyzed data from a particular year; for a sensible choice of class intervals, a histogram did not look at all normal, but after further investigations it was determined that this was due to some hospitals measuring weight in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this gave a histogram that was well described by a normal distribution.] a. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(4000 \mathrm{~g}\) ? Is between 3000 and \(4000 \mathrm{~g}\) ? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds \(7 \mathrm{lb}\) ? d. How would you characterize the most extreme .1\% of all birth weights? e. If \(X\) is a random variable with a normal distribution and \(a\) is a numerical constant \((a \neq 0\) ), then \(Y=a X\) also has a normal distribution. Use this to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from part (c). How does this compare to your previous answer?

In each case, determine the value of the constant \(c\) that makes the probability statement correct. a. \(\Phi(c)=.9838\) b. \(P(0 \leq Z \leq c)=.291\) c. \(P(c \leq Z)=.121\) d. \(P(-c \leq Z \leq c)=.668\) e. \(P(c \leq|Z|)=.016\)

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