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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 \(2 \mathrm{hr}\) 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. 0.9616 b. 0.0614

Step by step solution

01

Define the distributions

Let the velocity of the first airplane be denoted as \( V_1 \) and the velocity of the second airplane as \( V_2 \). \( V_1 \sim N(520, 10^2) \) and \( V_2 \sim N(500, 10^2) \) are both normally distributed random variables.
02

Calculate the relative displacement after 2 hours

The position of the first plane after 2 hours is \( S_1 = 10 + 2V_1 \) and the position of the second plane is \( S_2 = 2V_2 \). The relative displacement \( D \) between the planes after 2 hours is given by \( D = S_1 - S_2 = 10 + 2V_1 - 2V_2 = 10 + 2(V_1 - V_2) \). Define a new random variable for the velocity difference: \( X = V_1 - V_2 \). Since \( V_1 \) and \( V_2 \) are independent, \( X \sim N(520 - 500, 10^2 + 10^2) = N(20, 200) \). The displacement \( D = 10 + 2X \).
03

Solve part (a)

For part (a), we want the probability that the second plane has not caught up to the first plane, i.e., \( D > 0 \). The condition for the second plane to catch up is \( 10 + 2X > 0 \) or \( X > -5 \).Let \( Y \sim N(20, 200) \).Calculate the probability: \[ P(X > -5) = 1 - P(X \leq -5) \]. Convert to standard normal: \[ Z = \frac{X - 20}{\sqrt{200}} \rightarrow \frac{-5 - 20}{\sqrt{200}} \approx -1.77 \]. Using standard normal tables, \( P(Z \leq -1.77) \approx 0.0384 \), so:\[ P(Z > -1.77) = 0.9616 \].
04

Solve part (b)

For part (b), we calculate the probability that the planes are separated by at most 10 km: \( |D| \leq 10 \). This translates to \(-10 \leq 10 + 2X \leq 10 \). Solving for \( X \), we find: \(-10 - 10 \leq 2X \leq 0 \Rightarrow -10 \leq X \leq 0 \).Calculate \( P(-10 \leq X \leq 0) \). Convert endpoints to standard normal:\[ P\left(\frac{-10 - 20}{\sqrt{200}} \leq Z \leq \frac{0 - 20}{\sqrt{200}}\right) \Rightarrow P(-2.12 \leq Z \leq -1.41) \] Using standard normal tables, \( P(Z \leq -1.41) \approx 0.0793 \) and \( P(Z \leq -2.12) \approx 0.0179 \).Therefore,\[ P(-2.12 \leq Z \leq -1.41) = 0.0793 - 0.0179 = 0.0614 \].

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Key Concepts

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

Random Variables
In statistics, a **Random Variable** is a numerical outcome of a random phenomenon. It is essentially a variable that assumes different values based on the outcomes of a probabilistic experiment. For instance, in our airplane scenario, the velocities of the airplanes, denoted by \( V_1 \) and \( V_2 \), are examples of random variables. Each of these follows a Normal Distribution, one of the most widely used distributions for continuous random variables.
When dealing with random variables, we often aim to understand their behavior using a probability distribution. This distribution describes all the conceivable values a random variable can take and the likelihood of each of these.
  • In this problem, \( V_1 \sim N(520, 100) \) and \( V_2 \sim N(500, 100) \), which means they have means of 520 and 500, respectively, and have a variance of 100.
  • By understanding how these random variables behave, we can make predictions about the system they describe, such as predicting the future position of the airplanes after a given time.
Probability Calculation
The process of **Probability Calculation** involves using mathematical concepts to determine the likelihood of different outcomes of a random process. In the airplane problem, we need to calculate the probability that certain events occur after 2 hours:
- **Event 1:** The second plane has not caught up to the first plane, which means it is important to find the probability that the relative displacement is more than zero.- **Event 2:** The separation between the two planes is at most 10 km.To do this, we use the properties of the normal distribution of the random variables and form a new random variable for the velocity difference between the two planes. This velocity difference, \( X = V_1 - V_2 \), is also normally distributed because \( V_1 \) and \( V_2 \) are both normal random variables that are independent of each other.
Next, by using transformations (using the formula \( Z = \frac{X - ext{{mean}}}{ ext{{standard deviation}}} \)), we convert our calculated values to a standard normal distribution to find the probability values on standard normal tables. By interpreting these results, we can perform accurate probability calculations to solve different parts of the problem.
Standard Normal Table
The **Standard Normal Table**, also known as the Z-table, is an integral tool in statistical probability calculations. It helps determine the probability that a standard normal random variable is less than or equal to a certain value. This is essential when dealing with probabilities related to normal distributions.
When converting a normal random variable to a standard normal variable (denoted as \( Z \)), it allows us to utilize this table effectively. In the exercise, after calculating \( Z \)-values for specific scenarios, we refer to these tables to obtain probability values:
  • For example, in part (a), a \( Z \)-value of -1.77 was found, and using the table, the probability \( P(Z \leq -1.77) \) was around 0.0384.
  • For part (b), the Z-table helped find probabilities for \( Z \) between certain values, allowing us to predict if the separation would be at most 10 km.
Thus, the standard normal table is indispensable, offering precise probability values crucial for solving problems involving normal distributions.
Independent Events
In probability theory, **Independent Events** are those that do not affect each other's outcomes. When one event occurs, it does not influence the probability of the other occurring. This property of independence is vital in dealing with random variables, especially in the current scenario involving two airplanes.
Each airplane's speed is an independent event. This means that the speed of the first plane \( V_1 \) does not impact the probability distribution of the second plane's speed \( V_2 \), and vice versa. This independence is a key factor in forming the new random variable \( X = V_1 - V_2 \) for this exercise because it simplifies probability calculations:
  • Since \( V_1 \) and \( V_2 \) are independent, the variance of \( X \) is the sum of the variances of \( V_1 \) and \( V_2 \), doubling the variance to 200 in this case.
  • This independence simplifies our calculations since adding or subtracting independent normally distributed variables results in a normal distribution.
Understanding and identifying independence between events or variables enables more accurate probability predictions and is fundamental in solving complex probabilistic problems.

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