/*! 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 26 Suppose that \(10^{6}\) people a... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 26

Suppose that \(10^{6}\) people arrive at a service station at times that are independent random variables, each of which is uniformly distributed over \(\left(0,10^{6}\right)\). Let \(N\) denote the number that arrive in the first hour. Find an approximation for \(P\\{N=i\\}\)

Short Answer

Expert verified
In this problem, we can approximate the continuous uniform distribution using a Poisson distribution. The average number of people arriving in the first hour is found to be \( \lambda = 1 \). Using the Poisson distribution formula, \( P\{N=i\} = \frac{e^{-\lambda} \lambda^i}{i!} \), we can approximate the probability of any given number (i) of people arriving in the first hour.

Step by step solution

01

Determine the average number of arrivals in the first hour

We know that \( 10^6 \) people arrive uniformly distributed over the interval \( (0, 10^6) \). Since the number of people is equal to the length of the interval, and we are interested in the first hour (which is the interval \( (0,1) \)), we can compute the average number of arrivals in the first hour as follows: Average number of arrivals in the first hour, \( \lambda \) = \(\frac{\text{Length of the interval of interest}}{\text{Length of the total interval}} \times \text{Total number of arrivals}\) \( \lambda \) = \(\frac{1}{10^6} \times 10^6\) \( \lambda \) = 1
02

Apply the Poisson distribution formula

Now that we have the average number of arrivals in the first hour (\( \lambda = 1\)), we can use the Poisson distribution formula to approximate the probability \( P\{N=i\} \) for any given value of i: \( P\{N=i\} \) = \(\frac{e^{-\lambda} \lambda^i}{i!}\) Here, \( e \) is the base of the natural logarithm (approximately 2.71828), \( \lambda \) is the average number of arrivals in the first hour, and i is the number of arrivals we want to compute the probability for.
03

Compute the approximation for P{N=i}

Using the Poisson distribution formula, we can compute the approximation for \( P\{N=i\} \) for any given value of i. Remember that \( \lambda = 1 \) for this problem. For example, if we want to find the probability of 2 people arriving in the first hour, we'd plug in \( i = 2 \) to the formula: \( P\{N=2\} \) = \(\frac{e^{-1} \cdot 1^2}{2!} \) To find the probability of any other number of arrivals (i), we would plug in the desired value of i and compute the Poisson distribution formula accordingly.

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!

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

Let \(X_{1}, X_{2}, \ldots\) be a sequence of independent uniform \((0,1)\) random variables. For a fixed constant \(c\), define the random variable \(N\) by $$ N=\min \left\\{n: X_{n}>c\right\\} $$ Is \(N\) independent of \(X_{N}\) ? That is, does knowing the value of the first random variable that is greater than \(c\) affect the probability distribution of when this random variable occurs? Give an intuitive explanation for your answer.

An ambulance travels back and forth, at a constant speed, along a road of length \(L\). At a certain moment of time an accident occurs at a point uniformly distributed on the road. [That is, its distance from one of the fixed ends of the road is uniformly distributed over \((0, L)\).] Assuming that the ambulance's location at the moment of the accident is also uniformly distributed, compute, assuming independence, the distribution of its distance from the accident.

Let \(X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)}\) be the ordered values of \(n\) independent uniform \((0,1)\) random variables. Prove that for \(1 \leq k \leq n+1\), $$ P\left\\{X_{(k)}-X_{(k-1)}>t\right\\}=(1-t)^{n} $$ where \(X_{0} \equiv 0, X_{n+1} \equiv t\).

Consider a sequence of independent Bernoulli trials, each of which is a success with probability \(p\). Let \(X_{1}\) be the number of failures preceding the first success, and let \(X_{2}\) be the number of failures between the first two successes. Find the joint mass function of \(X_{1}\) and \(X_{2}\)

Each throw of a unfair die lands on each of the odd numbers \(1,3,5\) with probability \(C_{3}\) and on each of the even numbers with probability \(2 C\). (a) Find \(C\). (b) Suppose that the die is tossed. Let \(X\) equal 1 if the result is an even -number, and let it be 0 otherwise. Also, let \(Y\) equal 1 if the result is a number greater than three and let it be 0 otherwise. Find the joint probability mass function of \(X\) and \(Y\). Suppose now that 12 independent tosses of the die are made. e(c) Find the probability that each of the six outcomes occurs exactly twice. (d) Find the probability that 4 of the outcomes are either one or two, 4 are either three or four, and 4 are either five or six. (e) Find the probability that at least 8 of the tosses land on even numbers.
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Geometry

Read Explanation

Decision Maths

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.