/*! 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 68 Two airplanes are flying in the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

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 \((\mathrm{km} / \mathrm{hr})\) is normally distributed with mean 520 and standard deviation 10 and the second plane's speed is also normally distributed with mean and standard deviation 500 and 10 , respectively. a. What is the probability that after 2 hr 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 \(2 \mathrm{hr}\).

Short Answer

Expert verified
a. Probability is approximately 0.9616. b. Probability is approximately 0.0409.

Step by step solution

01

Define Variables

Let the speed of the first plane be denoted as \(X\) and that of the second plane as \(Y\). The given data specifies \(X \sim N(520, 10^2)\) and \(Y \sim N(500, 10^2)\). The relative position after 2 hours depends on the difference in distance traveled by the two planes, starting with a head start of 10 km for plane 1.
02

Express Conditions Mathematically

After 2 hours, the first plane travels \(2X\) km and the second plane \(2Y\) km. Thus, the distance apart would be \(D = 2X - 2Y + 10\). Define \(Z = X - Y\), then \(D = 2Z + 10\). To find the probability that the second plane has not caught up, find \(P(D > 0) = P(2Z + 10 > 0) = P(Z > -5)\). For part b, find \(P(0 \leq D \leq 10)\) or \(P(0 \leq 2Z + 10 \leq 10)\), simplifying to \(-5 \leq Z \leq 0\).
03

Calculate the Distribution of Z

Since \(Z = X - Y\), \(Z\) is normally distributed with mean \(\mu_Z = \mu_X - \mu_Y = 520 - 500 = 20\) and variance \(\sigma_Z^2 = \sigma_X^2 + \sigma_Y^2 = 10^2 + 10^2 = 200\). Therefore, \(Z \sim N(20, 200)\).
04

Compute Probability for Part a

To compute \(P(Z > -5)\), use the transformation \(P(Z > -5) = P\left(\frac{Z - 20}{\sqrt{200}} > \frac{-5 - 20}{\sqrt{200}}\right) = P\left(\frac{Z - 20}{14.14} > -1.77\right)\). Using the standard normal table, \(P(Z > -1.77) = 1 - P(Z < -1.77)\). Find the corresponding value in a standard normal distribution table to get approximately 0.9616.
05

Compute Probability for Part b

For the probability \(-5 \leq Z \leq 0\), transform similarly: \(P(-5 \leq Z \leq 0) = P\left(\frac{-5-20}{14.14} \leq \frac{Z-20}{14.14} \leq \frac{0-20}{14.14}\right) = P(-1.77 \leq Z' \leq -1.41)\), where \(Z'\) is standard normal. Using the table for normal distribution, find \(P(Z' < -1.41)\) and \(P(Z' < -1.77)\) and compute their difference: \(0.0793 - 0.0384 = 0.0409\).

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 91Ó°ÊÓ!

Key Concepts

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

Normal Distribution
The normal distribution is a fundamental concept in statistics and probability theory, used to describe how data values are distributed. It's often referred to as a "bell curve" because of its bell-shaped appearance when graphed. Most of the data points will cluster around the mean or average value, with fewer and fewer points as you move away from the mean on either side.

The normal distribution is defined by two parameters: the mean (\( \mu \)) and the standard deviation (\( \sigma \)). The mean is the center of the distribution, and the standard deviation measures the spread of the distribution. In our exercise, the speeds of both airplanes are normally distributed. The first plane has a mean speed of 520 km/hr, and the second plane has a mean speed of 500 km/hr, both with a standard deviation of 10 km/hr.

Understanding these parameters helps predict how likely airplanes or other objects will deviate from their average speeds, which is essential in practical scenarios like air traffic management.
Random Variables
In statistics, a random variable is a variable whose values are determined by the outcomes of a random process. Random variables can be discrete or continuous. For example, flipping a coin results in a discrete random variable, while measuring a person's height is a continuous random variable.

In this problem, the speed of each airplane is represented as a random variable, denoted as \(X\) for the first plane and \(Y\) for the second plane. These variables are continuous because the speed can take on any value within a given range, governed by the normal distribution parameters specified above.

Random variables enable us to calculate probabilities and make statistical inferences. In our scenario, examining \(X - Y\) gives us insights into the relative speeds of the two planes. The difference encapsulates whether one plane can catch up to another over a period.
Standard Normal Distribution
The standard normal distribution is a special case of the normal distribution where the mean is 0 and the standard deviation is 1. This type of distribution is particularly useful because any normal distribution can be transformed into a standard normal distribution using standardization.Standardizing converts a normal random variable \(Z\) with any mean \(\mu\) and standard deviation \(\sigma\) into a version \(Z'\) as follows:\[Z' = \frac{Z - \mu}{\sigma}\]In our problem, we calculate \(Z = X - Y\), which is normally distributed with mean 20 and variance 200. We then transform \(Z\) to \(Z'\) such that we can use the standard normal distribution tables to find probabilities more easily.

By doing this transformation, we compare \(Z\) values against the unified standard distribution values, allowing the use of pre-calculated tables to find probabilities quickly.
Probability Calculation
Probability calculation is a critical aspect of determining likelihoods in uncertain scenarios. In the given problem, we're tasked with finding probabilities related to the position and speeds of airplanes.Firstly, for part (a), we need to calculate the probability that the second plane does not overtake the first plane. Mathematically, this means finding the probability \( P(Z > -5) \). By converting to the standard normal distribution using the transformation, we find \( P(Z' > -1.77) \), which corresponds to probabilities given by standard normal tables.

Secondly, part (b) asks for the probability that the planes are separated by at most 10 km. This involves calculating probabilities within a certain range: \( -5 \leq Z \leq 0 \). This requires calculating \( P(-1.77 \leq Z' \leq -1.41) \) after converting to a standard normal variable. Here, \( P \) values are obtained by subtracting cumulative probabilities given in standard tables.

This process showcases how statistical tools convert real-world problems into mathematical expressions that can be computed to assist in decision-making.

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

A service station has both self-service and full-service islands. On each island, there is a single regular unleaded pump with two hoses. Let \(X\) denote the number of hoses being used on the self-service island at a particular time, and let \(Y\) denote the number of hoses on the full-service island in use at that time. The joint pmf of \(X\) and \(Y\) appears in the accompanying tabulation. \begin{tabular}{ll|ccc} \(p(x, y)\) & & 0 & 1 & 2 \\ \hline & 0 & \(.10\) & 04 & \(.02\) \\ \(x\) & 1 & \(.08\) & \(.20\) & \(.06\) \\ & 2 & \(.06\) & \(.14\) & \(.30\) \end{tabular} a. What is \(P(X=1\) and \(Y=1)\) ? b. Compute \(P(X \leq 1\) and \(Y \leq 1)\). c. Give a word description of the event \(\\{X \neq 0\) and \(Y \neq 0\\}\), and compute the probability of this event. d. Compute the marginal pmf of \(X\) and of \(Y\). Using \(p_{X}(x)\), what is \(P(X \leq 1)\) ? e. Are \(X\) and \(Y\) independent rv's? Explain.

Carry out a simulation experiment using a statistical computer package or other software to study the sampling distribution of \(\bar{X}\) 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 1000 replications. For which of these sample sizes does the \(\bar{X}\) sampling distribution appear to be approximately normal?

It is known that \(80 \%\) of all brand A zip drives work in a satisfactory manner throughout the warranty period (are "successes"). Suppose that \(n=10\) drives are randomly selected. Let \(X=\) the number of successes in the sample. The statistic \(X / n\) is the sample proportion (fraction) of successes. Obtain the sampling distribution of this statistic. [Hint: One possible value of \(X / n\) is \(.3\), corresponding to \(X=3\). What is the probability of this value (what kind of random variable is \(X\) )?]

You have two lightbulbs for a particular lamp. Let \(X=\) the lifetime of the first bulb and \(Y=\) the lifetime of the second bulb (both in 1000 s of hours). Suppose that \(X\) and \(Y\) are independent and that each has an exponential distribution with parameter \(\lambda=1\). a. What is the joint pdf of \(X\) and \(Y\) ? b. What is the probability that each bulb lasts at most 1000 hours (i.e., \(X \leq 1\) and \(Y \leq 1\) )? c. What is the probability that the total lifetime of the two bulbs is at most 2 ? [Hint: Draw a picture of the region \(A=\\{(x, y): x \geq 0, y \geq 0, x+y \leq 2\\}\) before integrating.] d. What is the probability that the total lifetime is between 1 and 2 ?

Suppose the distribution of the time \(X\) (in hours) spent by students at a certain university on a particular project is gamma with parameters \(\alpha=50\) and \(\beta=2\). Because \(\alpha\) is large, it can be shown that \(X\) has approximately a normal distribution. Use this fact to compute the approximate probability that a randomly selected student spends at most 125 hours on the project.

See all solutions

Recommended explanations on Math Textbooks

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.