/*! 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 24 Let \(x\) denote the amount of g... [FREE SOLUTION] | 91Ó°ÊÓ

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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, \(P(x >= 1/4) = 1 - P(x < 1/4) = 1 - 3/32 = 29/32\).

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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.

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Most popular questions from this chapter

Suppose that fuel efficiency for a particular model car under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this type is between 29 and \(31 \mathrm{mpg}\) ? b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg}\) ? c. If three cars of this model are randomly selected, what is the probability that all three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(c\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(c\) (i.e., \(P(x>c)=.95\) ).

Accurate labeling of packaged meat is difficult because of weight decrease resulting from moisture loss (defined as a percentage of the package's original net weight). Suppose that moisture loss for a package of chicken breasts is normally distributed with mean value \(4.0 \%\) and standard deviation \(1.0 \% .\) (This model is suggested in the paper "Drained Weight Labeling for Meat and Poultry: An Economic Analysis of a Regulatory Proposal," Journal of Consumer Affairs [1980]: 307-325.) Let \(x\) denote the moisture loss for a randomly selected package. a. What is the probability that \(x\) is between \(3.0 \%\) and \(5.0 \%\) ? b. What is the probability that \(x\) is at most \(4.0 \%\) ? c. What is the probability that \(x\) is at least \(7.0 \%\) ? d. Find a number \(z^{*}\) such that \(90 \%\) of all packages have moisture losses below \(z^{*} \%\). e. What is the probability that moisture loss differs from the mean value by at least \(1 \%\) ?

A local television station sells \(15-\mathrm{sec}, 30-\mathrm{sec}\), and 60 -sec advertising spots. Let \(x\) denote the length of a randomly selected commercial appearing on this station, and suppose that the probability distribution of \(x\) is given by the following table: $$ \begin{array}{lrrr} x & 15 & 30 & 60 \\ p(x) & .1 & .3 & .6 \end{array} $$ a. Find the average length for commercials appearing on this station. b. If a 15 -sec spot sells for $$\$ 500$$, a 30 -sec spot for $$\$ 800$$, and a 60 -sec spot for $$\$ 1000$$, find the average amount paid for commercials appearing on this station. (Hint: Consider a new variable, \(y=\) cost, and then find the probability distribution and mean value of \(y .\) )

The Los Angeles Times (December 13,1992 ) reported that what airline passengers like to do most on long flights is rest or sleep; in a survey of 3697 passengers, almost \(80 \%\) did so. Suppose that for a particular route the actual percentage is exactly \(80 \%\), and consider randomly selecting six passengers. Then \(x\), the number among the selected six who rested or slept, is a binomial random variable with \(n=6\) and \(\pi=.8\). a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6)\), the probability that all six selected passengers rested or slept. c. Determine \(P(x \geq 4)\).

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?

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