Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Let \(x\) denote the amount of gravel sold (in tons) during a randomly selected week at a particular sales facility. Suppose that the density curve has height \(f(x)\) above the value \(x\), where $$ f(x)=\left\\{\begin{array}{ll} 2(1-x) & 0 \leq x \leq 1 \\ 0 & \text { otherwise } \end{array}\right. $$ The density curve (the graph of \(f(x)\) ) is shown in the following figure: Use the fact that the area of a triangle \(=\frac{1}{2}\) (base)(height) to calculate each of the following probabilities: a. \(P\left(x<\frac{1}{2}\right)\) b. \(P\left(x \leq \frac{1}{2}\right)\) c. \(P\left(x<\frac{1}{4}\right)\) d. \(P\left(\frac{1}{4}

Short Answer

Expert verified
The probabilities asked for in the problem are: a. 1/8, b. 1/8, c. 3/32, d. 1/32, e. 7/8, f. 29/32.

Step by step solution

01

Calculate P(x < 1/2)

The probability \(P(x < 1/2)\) is the area of a triangle with base \(x = 1/2\) and height \(f(x) = 2(1 - 1/2) = 1\). Hence, \(P(x < 1/2) = (1/2) * 1/2 * 1 = 1/8\).
02

Calculate P(x

The probability \(P(x <= 1/2)\) is equal to the probability \(P(x < 1/2)\) because for continuous random variables, the probabilities at individual points are zero. Thus, \(P(x <= 1/2) = 1/8\).
03

Calculate P(x < 1/4)

The probability \(P(x < 1/4)\) is the area of a triangle with base \(x = 1/4\) and height \(f(x) = 2(1 - 1/4) = 3/2\). Hence, \(P(x < 1/4) = (1/2) * 1/4 * 3/2 = 3/32\).
04

Calculate P(1/4 < x < 1/2)

The probability \(P(1/4 < x < 1/2)\) is the difference between the areas of two triangles: one with base \(x = 1/2\) (calculated in Step 1), and another with base \(x = 1/4\) (calculated in Step 3). Thus, \(P(1/4 < x < 1/2) = P(x < 1/2) - P(x < 1/4) = 1/8 - 3/32 = 1/32\).
05

Calculate the probability that gravel sold exceeds 1/2 ton

The probability that gravel sold exceeds 1/2 ton is equal to 1 minus the probability that it is less than or equal to 1/2 ton. Thus, \(P(x > 1/2) = 1 - P(x <= 1/2) = 1 - 1/8 = 7/8\).
06

Calculate the probability that gravel sold is at least 1/4 ton

The probability that gravel sold is at least 1/4 ton is equal to 1 minus the probability that it is less than 1/4 ton. Thus, \(P(x >= 1/4) = 1 - P(x < 1/4) = 1 - 3/32 = 29/32\).

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 Vaia!

Key Concepts

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

Continuous Random Variables
When we talk about continuous random variables, we refer to quantities that can take an infinite number of possible values within a given range. For example, the amount of gravel sold, measured in tons, is a continuous random variable because it can potentially be any real number within the possible selling capacity. Unlike discrete random variables, which have countable values (like the number of textbooks on a shelf), continuous random variables require an associated probability density function (PDF) to describe the likelihood of the variable being in a specific range. The key feature distinguishing continuous variables is that the probability of it assuming any single, exact value is zero; instead, we focus on intervals to determine probabilities.

For instance, if we consider the exercise at hand, the random variable 'x,' representing gravel sold, is treated as a continuous random variable since x can take on any value within the range (0, 1). Probabilities are computed using the PDF, not by counting individual outcomes as you would with discrete variables.
Area under the Curve
The probability density function (PDF) for a continuous random variable is graphically represented by a curve on a graph. The probability of the variable falling within a certain range is determined by the area under the curve between two points. Because of this, understanding the geometry behind the curve becomes necessary. The total area under the curve for a properly defined PDF always equals 1, signifying a 100% probability that the variable will take on a value within the given range.

In the exercise, the PDF of the random variable 'x' is defined by the function \(f(x)\) and considering the curve is a straight line forming a triangle in this case, calculating areas is straightforward using the familiar formula for the area of a triangle. It's important to realize that calculating probabilities involves finding the area of various shapes under the curve, where the x-axis represents the possible values and the y-axis shows the density.
Probability Calculations
In probability theory, especially for continuous random variables, it is often necessary to calculate the likelihood, or probability, that the variable falls within a certain range. These calculations are derived from the properties of the PDF and the corresponding graphical area under the curve. To execute these calculations correctly, one must analyze the shape of the graph and use the appropriate geometric formulas.

Referring back to our gravel example 'x', probabilities such as \( P(x < 1/2) \) or \( P(x > 1/2) \) are deduced by considering the shape under the curve—triangles in this case—and applying the area formula. Thus, taking the appropriate base and height from the function \(f(x)\) and implementing the formula for the area of a triangle provides the solution to various probability inquiries detailed in the exercise. Learning to visualize and calculate the area is a critical skill for effective probability analysis with continuous random variables.

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 \(x\) have a binomial distribution with \(n=50\) and \(\pi=.6\), so that \(\mu=n \pi=30\) and \(\sigma=\sqrt{n \pi(1-\pi)}=\) 3.4641. Calculate the following probabilities using the normal approximation with the continuity correction: a. \(P(x=30)\) b. \(P(x=25)\) c. \(P(x \leq 25)\) d. \(P(25 \leq x \leq 40)\) e. \(P(25

A coin is spun 25 times. Let \(x\) be the number of spins that result in heads (H). Consider the following rule for deciding whether or not the coin is fair: Judge the coin fair if \(8 \leq x \leq 17\). Judge the coin biased if either \(x \leq 7\) or \(x \geq 18\). a. What is the probability of judging the coin biased when it is actually fair? b. What is the probability of judging the coin fair when \(P(\mathrm{H})=.9\), so that there is a substantial bias? Repeat for \(P(\mathrm{H})=.1 .\) c. What is the probability of judging the coin fair when \(P(\mathrm{H})=.6\) ? when \(P(\mathrm{H})=.4 ?\) Why are the probabilities so large compared to the probabilities in Part (b)? d. What happens to the "error probabilities" of Parts (a) and (b) if the decision rule is changed so that the coin is judged fair if \(7 \leq x \leq 18\) and unfair otherwise? Is this a better rule than the one first proposed? Explain.

Suppose that the \(\mathrm{pH}\) of soil samples taken from a certain geographic region is normally distributed with a mean \(\mathrm{pH}\) of \(6.00\) and a standard deviation of \(0.10 .\) If the \(\mathrm{pH}\) of a randomly selected soil sample from this region is determined, answer the following questions about it: a. What is the probability that the resulting \(\mathrm{pH}\) is between \(5.90\) and \(6.15 ?\) b. What is the probability that the resulting \(\mathrm{pH}\) exceeds \(6.10 ?\) c. What is the probability that the resulting \(\mathrm{pH}\) is at most 5.95? d. What value will be exceeded by only \(5 \%\) of all such pH values?

A pizza company advertises that it puts \(0.5 \mathrm{lb}\) of real mozzarella cheese on its medium pizzas. In fact, the amount of cheese on a randomly selected medium pizza is normally distributed with a mean value of \(0.5 \mathrm{lb}\) and \(\mathrm{a}\) standard deviation of \(0.025 \mathrm{lb}\). a. What is the probability that the amount of cheese on a medium pizza is between \(0.525\) and \(0.550 \mathrm{lb}\) ? b. What is the probability that the amount of cheese on a medium pizza exceeds the mean value by more than 2 standard deviations? c. What is the probability that three randomly selected medium pizzas all have at least \(0.475 \mathrm{lb}\) of cheese?

State whether each of the following random variables is discrete or continuous: a. The number of defective tires on a car b. The body temperature of a hospital patient c. The number of pages in a book d. The number of draws (with replacement) from a deck of cards until a heart is selected e. The lifetime of a lightbulb

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.

Sign-up for free