/*! 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 37 Two airplanes are flying in the ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 37

Two airplanes are flying in the same direction in adjacent parallel corridors. At time \(t=0\), the first airplane is \(10 \mathrm{~km}\) ahead of the second one. Suppose the speed of the first plane \((\mathrm{km} / \mathrm{h})\) is normally distributed with mean 520 and standard deviation 10 and the second plane's speed, independent of the first, is also normally distributed with mean and standard deviation 500 and 10 , respectively. a. What is the probability that after \(2 \mathrm{~h}\) of flying, the second plane has not caught up to the first plane? b. Determine the probability that the planes are separated by at most \(10 \mathrm{~km}\) after \(2 \mathrm{~h}\).

Short Answer

Expert verified
a. The probability is 0.0793. b. The probability is 0.2604.

Step by step solution

01

Setup the Speeds

Let the speed of the first plane be denoted by the random variable \(X \sim N(520, 10^2)\), and the speed of the second plane by \(Y \sim N(500, 10^2)\).
02

Determine Distance Covered

The distance covered by the first plane after 2 hours is given by \(D_1 = 520 \times 2 + A\), where \(A\) is a normally distributed random variable with mean 0 and standard deviation \(20\) (since the speed deviation is \(20\) over 2 hours). Similarly, the distance covered by the second plane is \(D_2 = 500 \times 2 + B\), where \(B\) is also normally distributed with mean 0 and standard deviation \(20\).
03

Formulate the Problem

The second airplane catching up to the first is described by the event \(D_2 \geq D_1 - 10\). After 2 hours, \(D_1 - D_2 = 20 + A - B\). We want \(P(A - B < 0)\) for part (a) and \(P(0 < A - B < 10)\) for part (b).
04

Calculate the Relative Speed Distribution

Define \(Z = A - B\), where \(Z\) is also normally distributed with mean 0 (since \(E[A - B] = 0 - 0 \)) and variance \(10^2 + 10^2 = 200\), so \(Z \sim N(0, 200)\).
05

Solve Part (a) - Probability that Second Plane has not Caught Up

We calculate \(P(Z < -20)\). Standardize \(Z\) using the transformation \(Z^* = \frac{Z}{\sqrt{200}} \sim N(0,1)\). Then \(P(Z < -20) = P\left(Z^* < \frac{-20}{\sqrt{200}}\right) = P(Z^* < -1.414)\). Using the standard normal distribution table, \(P(Z^* < -1.414) \approx 0.0793\).
06

Solve Part (b) - Probability Planes are Separated by At Most 10 km

We find \(P(0 < Z < 10)\), which is \(P\left(0 < Z^* < \frac{10}{\sqrt{200}}\right)\). For \(Z^* = 0.707\), \(P(Z^* < 0.707) \approx 0.7604\), and \(P(Z^* < 0) = 0.5\). Hence, \(P(0 < Z^* < 0.707) = 0.7604 - 0.5 = 0.2604\).

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.

Random Variables
In probability and statistics, random variables are fundamental concepts used to describe numerical outcomes of random phenomena. In this exercise, both airplanes' speeds are represented by random variables:
  • The speed of the first airplane is described by the random variable \(X\), which follows a normal distribution with mean 520 km/h and standard deviation \(10\).
  • The speed of the second airplane is represented by the random variable \(Y\), which also follows a normal distribution, but with mean 500 km/h and standard deviation \(10\).
These distributions imply that while the speeds are centered around their means, they can vary, reflecting potential speed fluctuations of each plane. The distribution tells us how likely these fluctuations are.
Random variables are crucial because they allow us to make probabilistic statements about the speeds over time, helping us answer questions like the likelihood of one plane catching up to another.
Probability Calculation
Probability calculation involves determining the likelihood of certain outcomes. In this problem, we focus on two specific scenarios:
  • The probability that the second plane has not caught up with the first one after 2 hours of flying.
  • The probability that the planes are separated by at most 10 km after 2 hours.
To calculate these probabilities, we first determine the relative distances covered by each plane over time:
  • Distance by the first plane is affected by its random variable representing speed deviations, hence modeled by \(D_1 = 1040 + A\), where \(A\) is normally distributed with a mean of 0.
  • Similarly, for the second plane, it's \(D_2 = 1000 + B\), with \(B\) also normally distributed around 0.
The key here is the event described by \(Z = A - B\), the difference in these deviations which follows a normal distribution \(N(0, 200)\). Based on this, we can standardize and use statistical tables or calculators to find exact probabilities for the given scenarios.
Standardization of Normal Variables
Standardization is a method used to transform a normal random variable to a standard normal variable, making it easier to calculate probabilities. When a variable \(Z\) is normal with mean \(0\) and variance \(200\), we standardize it:\[Z^* = \frac{Z}{\sqrt{200}}\]This transformation converts \(Z\) into a standard normal distribution \(N(0, 1)\), which is key to accessing cumulative probability values on a standard normal distribution table.
In our problem:
  • To find the probability of the second plane not catching up, we standardize and calculate \(P(Z^* < -1.414)\).
  • For planes separated by at most 10 km, we calculate \(P(0 < Z^* < 0.707)\).
These transformations leverage the properties of the normal distribution, allowing us to use standardized tables, streamlining the calculation process.

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 Central Limit Theorem says that \(X\) is approximately normal if the sample size is large. More specifically, the theorem states that the standardized \(\bar{X}\) has a limiting standard normal distribution. That is, \((X-\mu) /(\sigma / \sqrt{n})\) has a distribution approaching the standard normal. Can you reconcile this with the Law of Large Numbers? If the standardized \(X\) is approximately standard normal, then what about \(\bar{X}\) itself?

Let \(X\) represent the amount of gasoline (gallons) purchased by a randomly selected customer at a gas station. Suppose that the mean value and standard deviation of \(X\) are \(11.5\) and \(4.0\), respectively. a. In a sample of 50 randomly selected customers, what is the approximate probability that the sample mean amount purchased is at least 12 gallons? b. In a sample of 50 randomly selected customers, what is the approximate probability that the total amount of gasoline purchased is at most 600 gallons. c. What is the approximate value of the 95 th percentile for the total amount purchased by 50 randomly selected customers.

A shipping company handles containers in three different sizes: (1) \(27 \mathrm{ft}^{3}(3 \times 3 \times 3),(2) 125 \mathrm{ft}^{3}\), and \((3) 512 \mathrm{ft}^{3}\). Let \(X_{i}(i=1,2,3)\) denote the number of type \(i\) containers shipped during a given week. With \(\mu_{i}=E\left(X_{j}\right)\) and \(\sigma_{i}^{2}=V\left(X_{i}\right)\), suppose that the mean values and standard deviations are as follows: $$ \begin{array}{lll} \mu_{1}=200 & \mu_{2}=250 & \mu_{3}=100 \\ \sigma_{1}=10 & \sigma_{2}=12 & \sigma_{3}=8 \end{array} $$ a. Assuming that \(X_{1}, X_{2}, X_{3}\) are independent, calculate the expected value and variance of the total volume shipped. [Hint: Volume = \(27 X_{1}+125 X_{2}+512 X_{3}\).] b. Would your calculations necessarily be correct if the \(X_{i}\) 's were not independent? Explain. c. Suppose that the \(X_{i}\) 's are independent with each one having a normal distribution. What is the probability that the total volume shipped is at most \(100,000 \mathrm{ft}^{3}\) ?

The number of parking tickets issued in a certain city on any given weekday has a Poisson distribution with parameter \(\lambda=50\). What is the approximate probability that a. Between 35 and 70 tickets are given out on a particular day? [Hint: When \(\lambda\) is large, a Poisson rv has approximately a normal distribution.] b. The total number of tickets given out during a 5 -day week is between 225 and 275 ?

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \(X\) when the population distribution is lognormal with \(E[\ln (X)]=3\) and \(V[\ln (X)]=1\). Consider the four sample sizes \(n=10,20,30\), and 50 , and in each case use 500 replications. For which of these sample sizes does the \(X\) sampling distribution appear to be approximately normal?
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation

Mechanics Maths

Read Explanation

Calculus

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.