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Chapter 3: Problem 44

Each rear tire on an experimental airplane is supposed to be filled to a pressure of 40 pound per square inch (psi). Let \(X\) denote the actual air pressure for the right tire and \(Y\) denote the actual air pressure for the left tire. Suppose that \(X\) and \(Y\) are random variables with the joint density $$ f(x, y)=\left\\{\begin{array}{ll} k\left(x^{2}+y^{2}\right), & 30 \leq x<50; \\ & 30 \leq y<50; \\ 0, & \text { elsewhere. } \end{array}\right. $$ (a) Find \(k\). (b) Find \(\mathrm{P}(30 \leq X \leq 40\) and \(40 \leq Y<50)\) (c) Find the probability that both tires are underfilled.

Short Answer

Expert verified
The value of \(k\) is \(1/8000\). The probability that \(X\) is between 30 and 40, and \(Y\) is between 40 and 50 is \(7/60\). The probability that both tires are underfilled (both \(X\) and \(Y\) are between 30 and 40) is \(1/15\).

Step by step solution

01

Finding 'k' via double integration

To find the value of \(k\), integrate \(f(x, y)\) over the entire area (the range where f(x, y) is not zero) and set it equal to 1, as the total probability must sum up to 1. We are taking integral from 30 to 50 as per the function's definition and setting it equal to 1, like this: \[ 1 = \int_{30}^{50} \int_{30}^{50} k(x^2+y^2) dx dy \] This simplifies to \( k = 1/8000 \)
02

Find Probability for \(30 \leq X \leq 40\) and \(40 \leq Y < 50\)

Now we need to find the probability that X is between 30 and 40, and Y is between 40 and 50. We integrate \(f(x, y)\) within these limits after substituting 'k' from step 1: \[ P(30 \leq X \leq 40, 40 \leq Y < 50) = \int_{30}^{40} \int_{40}^{50} (1/8000)(x^2+y^2) dx dy \] Doing the integrals, we get the probability as \(7/60\).
03

Calculate the Probability of Tires Being Underfilled

Tires are underfilled when the pressure is below 40 psi. We need to find the joint probability that both X and Y are between 30 and 40. Integrate \(f(x, y)\) with these limits, after substituting 'k' from step 1: \[ P(30 \leq X, Y \leq 40) = \int_{30}^{40} \int_{30}^{40} (1/8000)(x^2+y^2) dx dy \] Doing the integrals, we find the probability as \(1/15\).

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Key Concepts

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

Double Integration
When we talk about double integration in the context of probability, we're referring to a mathematical process used to compute the probability of certain outcomes for two continuous random variables. Double integration essentially involves integrating a function of two variables over a certain range, which in probability terms, means finding the cumulative probability that two variables fall within a specified range.

For instance, let鈥檚 consider the experimental airplane tire pressure problem. To find the constant value 'k' that ensures our joint probability density function sums up to one (which all probability distributions must), we use double integration over the range of pressures where the density function is non-zero, from 30 to 50 psi for both the left and right tires. This process is akin to summing up all the infinitesimally small probabilities across the entire range of values that X and Y can take, guaranteeing that we account for every possible outcome.
Random Variables
Random variables are key elements in probability and statistics, representing a set of possible values from a random experiment. In our airplane tire scenario, the actual air pressures of the tires, denoted by the random variables 'X' for the right tire and 'Y' for the left tire, can be any value with a probabilistic outcome within the specified range.

These variables are 鈥榗ontinuous鈥� because they can take on any value within an interval (such as any pressure level between 30 psi and 50 psi). Understanding random variables and their distributions allows us to predict and analyze outcomes using mathematical models. The air pressure in both tires can be described using a joint probability density function, a concept that bridges the gap between random variables and their probabilities.
Probability Distribution
A probability distribution is a mathematical function that provides the probabilities of occurrence of different possible outcomes of an experiment. It describes how the probabilities are distributed over the values of the random variable. For continuous variables, we refer to this as a probability density function (PDF).

The joint probability density function, specifically, relates to two random variables and defines the probability that each variable falls within a certain range simultaneously. In the tire pressure case, we use the joint density function to describe how the pressures in the two tires are related in terms of probability. By finding the value of 'k' using double integration, we ensure that the total area under this joint PDF is equal to 1, as it must represent all possible outcomes.
Underfilled Tire Pressure
In a practical situation like ensuring the correct tire pressure of airplane tires, 'underfilled tire pressure' is a critical factor to consider for safety and performance. In the given exercise, a tire is considered underfilled if its air pressure is below 40 psi.

The problem asks us to calculate the probability that both tires are underfilled. To do this, we set the range of interest between 30 and 40 psi and apply double integration within these limits on the joint PDF. This offers us the probability that both 'X' and 'Y', representing the right and left tire pressures, fall below the threshold for being suitably filled. Understanding the implications of underfilled tires and how to calculate the associated risk is essential for applications ranging from practical engineering to quality control.

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Most popular questions from this chapter

Classify the following random variables as discrete or continuous: \(X\) : the number of automobile accidents per year in Virginia. \(Y:\) the length of time to play 18 holes of golf. \(M\) : the amount of milk produced yearly by a particular cow. \(N:\) the number of eggs laid each month by a hen. \(P:\) the number of building permits issued each month in a certain city. \(Q:\) the weight of grain produced per acre.

Let \(X\) denote the number of times a certain numerical control machine will malfunction: \(1,2,\) or 3 times on any given day. Let \(Y\) denote the number of times a technician is called on an emergency call. Their joint probability distribution is given as $$ \begin{array}{cc|ccc} & & & x & \\ {f(x, y)}& & 1 & 2 & 3 \\ \hline & 1 & 0.05 & 0.05 & 0.1 \\ \text { y } & 2 & 0.05 & 0.1 & 0.35 \\ & 3 & 0 & 0.2 & 0.1 \end{array} $$ (a) Evaluate the marginal distribution of \(X\). (b) Evaluate the marginal distribution of \(Y\). (c) Find \(P(Y=3 \mid X=2).\)

Suppose it is known from large amounts of historical data that \(X\), the number of cars that arrive at a specific intersection during a 20 second time period, is characterized by the following discrete probability function $$ /(x)=e^{-6} \frac{b^{x}}{x !}, \quad x=0,1.2, \ldots $$ (a) Find the probability that in a specific 20 -second time period, more than 8 cars arrive at the intersection. (b) Find the probabilitythat only 2 cars arrive.

Suppose a special type of small data processing firm is so specialized that some have difficulty making a profit in their first year of operation. The pdf that characterizes the proportion \(Y\) that make a profit is given by $$ f(x)=\left\\{\begin{array}{ll} k y^{4}(1-y)^{3}+ & 0 \leq y \leq 1, \\ 0_{1} & \text { elsewhere. } \end{array}\right. $$ (a) What is the value of \(k\) that renders the above a valid density function? (b) Find the probability that at most \(50 \%\) of the firms make a profit in the first year. (c) Find the probability that at least \(80 \%\) of the firms make a profit in the first year.

Find the probability distribution for the number of jazz CDs when 4 CDs are selected at random from a collection consisting of 5 jazz CDs, 2 classical CDs, and 3 rock CDs. Express your results by means of a formula.
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