/*! 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 13 You arrive at a bus stop at 10 A... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 13

You arrive at a bus stop at 10 A.M., knowing that the bus will arrive at some time uniformly distributed between 10 and 10: 30 (a) What is the probability that you will have to wait longer than 10 minutes? (b) If, at \(10: 15,\) the bus has not yet arrived, what is the probability that you will have to wait at least an additional 10 minutes?

Short Answer

Expert verified
In summary, the probability that you will have to wait longer than 10 minutes is \(\frac{2}{3}\), and the probability that you will have to wait at least an additional 10 minutes, given that the bus has not yet arrived at 10:15 AM, is \(\frac{1}{3}\).

Step by step solution

01

(a) Determine the probability density function

Firstly, we need to find the probability density function (pdf) of the bus arrival time. Since the bus will arrive at some time uniformly distributed between 10:00 AM and 10:30 AM, we can write the interval as [0, 30] minutes. The pdf of a uniform distribution is given by: \(f(x)=\frac{1}{b-a}\) for \(a \le x \le b\) In our case, \(a=0\) and \(b=30\). So, the pdf is: \(f(x)=\frac{1}{30-0}=\frac{1}{30}\) for \(0 \le x \le 30\)
02

(a) Calculate the probability of waiting longer than 10 minutes

To find the probability of waiting longer than 10 minutes, we need to calculate the area under the pdf curve from 10 minutes to 30 minutes. We can integrate the pdf function over this interval: \(P(X>10)=\int_{10}^{30} f(x) dx = \int_{10}^{30} \frac{1}{30} dx\) Now, we can solve this integral: \(P(X>10) = \left[ \frac{x}{30} \right]_{10}^{30} = \frac{30}{30} - \frac{10}{30} = \frac{2}{3}\) So, the probability that you will have to wait longer than 10 minutes is \(\frac{2}{3}\).
03

(b) Calculate the conditional probability

In this part of the exercise, we are asked to calculate the probability that you will have to wait at least an additional 10 minutes given that the bus has not arrived at 10:15 AM. This is a conditional probability problem. We can represent this probability as \(P(X>25|X>15)\). Using the properties of conditional probability, we can write: \(P(X>25|X>15)=\frac{P(X>25 \cap X>15)}{P(X>15)}\) Since \(X>25\) implies \(X>15\), the probability in the numerator is simply the probability of \(X>25\): \(P(X>25)=\int_{25}^{30} f(x) dx = \int_{25}^{30} \frac{1}{30} dx\) Now, we can solve this integral: \(P(X>25) = \left[ \frac{x}{30} \right]_{25}^{30} = \frac{30}{30} - \frac{25}{30} = \frac{1}{6}\) Next, we need to find the probability of the bus not arriving until after 10:15 AM: \(P(X>15)=\int_{15}^{30} f(x) dx = \int_{15}^{30} \frac{1}{30} dx\) Solve the integral: \(P(X>15) = \left[ \frac{x}{30} \right]_{15}^{30} = \frac{30}{30} - \frac{15}{30} = \frac{1}{2}\) Now, we can find the conditional probability: \(P(X>25|X>15)=\frac{P(X>25)}{P(X>15)}=\frac{\frac{1}{6}}{\frac{1}{2}}=\frac{1/6}{1/2}=\frac{2}{6}=\frac{1}{3}\) Thus, the probability that you will have to wait at least an additional 10 minutes, given that the bus has not yet arrived at 10:15 AM, is \(\frac{1}{3}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Probability Density Function
Understanding probability requires familiarity with the concept of a probability density function (pdf), which is crucial when we deal with continuous probability distributions. The pdf helps determine the likelihood that a random variable takes on a value within a specific interval.

For a continuous random variable, the probability density function represents how the total probability is distributed across possible values the variable can assume. By definition, it must satisfy two conditions: first, the pdf needs to be non-negative at each point; and second, when integrated over all possible values, it must equal to one, indicating a certainty that the random variable will take on a value within the given range.

A uniform distribution's pdf is particularly straightforward鈥攊t has the same value across the entire interval where the variable can take on values, and zero otherwise. The pdf for a uniform distribution is given by \( f(x) = \frac{1}{b-a} \) for \( a \le x \le b \), where the interval \[a, b\] contains all the possible values of the random variable. The simplicity of a uniform distribution's pdf reflects the concept that every outcome within the specified range is equally likely.
Conditional Probability
Another key concept in understanding probabilities in various scenarios is conditional probability. It's the probability of an event occurring given that another event has already occurred, which can have a significant impact on the outcome.

The conditional probability of an event B given an event A is denoted by \( P(B|A) \), and is calculated by the formula \( P(B|A) = \frac{P(A \cap B)}{P(A)} \) as long as \( P(A) > 0\). Essentially, it tells us how to adjust our expectations for B in light of the fact that A has happened.

Conditional probability can change our understanding of a situation quite dramatically. For example, in the context of waiting for a bus, if you know the bus hasn't arrived in the first 15 minutes, this knowledge updates the likelihood of how much longer you'll need to wait. If we consider the probability that two events occur together, compared to the probability that one occurs on its own, conditional probability provides a framework to quantify this relationship.
Uniform Distribution
In probability and statistics, a uniform distribution is one of the simplest forms of distributions where all outcomes are equally likely. If a random variable X follows a uniform distribution over an interval \[a, b\], any value within the interval is as likely as any other.

This distribution is characterized by two parameters, the minimum and maximum values \(a\) and \(b\), which define the range of possible outcomes. The uniform distribution is often represented graphically as a rectangle because the probability is perfectly evenly distributed over the interval, hence its alternate name, rectangular distribution.

In real-world applications, uniform distributions are ideal for representing scenarios where a random variable has a finite number of outcomes that are equally probable, such as drawing a random card from a well-shuffled deck or, as our exercise describes, waiting for a bus that is equally likely to arrive at any minute within a half-hour window.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The salaries of physicians in a certain speciality are approximately normally distributed. If 25 percent of these physicians earn less than \(\$ 180,000\) and 25 percent earn more than \(\$ 320,000,\) approximately what fraction earn (a) less than \(\$ 200,000 ?\) (b) between \(\$ 280,000\) and \(\$ 320,000 ?\)

Every day Jo practices her tennis serve by continually serving until she has had a total of 50 successful serves. If each of her serves is, independently of previous ones, successful with probability \(.4,\) approximately what is the probability that she will need more than 100 serves to accomplish her goal? Hint: Imagine even if Jo is successful that she continues to serve until she has served exactly 100 times. What must be true about her first 100 serves if she is to reach her goal?

Let \(Y\) be a lognormal random variable (see Example 7e for its definition) and let \(c>0\) be a constant. Answer true or false to the following, and then give an explanation for your answer. (a) \(c Y\) is lognormal; (b) \(c+Y\) is lognormal.

The number of years a radio functions is exponentially distributed with parameter \(\lambda=\frac{1}{8} .\) If Jones buys a used radio, what is the probability that it will be working after an additional 8 years?

A model for the movement of a stock supposes that if the present price of the stock is \(s,\) then after one period, it will be either \(u s\) with probability \(p\) or \(d s\) with probability \(1-p .\) Assuming that successive movements are independent, approximate the probability that the stock's price will be up at least 30 percent after the next 1000 periods if \(u=1.012, d=0.990,\) and \(p=.52\).
See all solutions

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.