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The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research [1984]: \(1169-1174\) ) suggests the uniform distribution on the interval from 7.5 to 20 as a model for \(x=\) depth (in centimeters) of the bioturbation layer in sediment for a certain region. a. Draw the density curve for \(x\). b. What is the height of the density curve? c. What is the probability that \(x\) is at most \(12 ?\) d. What is the probability that \(x\) is between 10 and 15 ? Between 12 and 17 ? Why are these two probabilities equal?

Short Answer

Expert verified
The density curve is a rectangle from 7.5 to 20 on the x-axis with a height of 0.08. The probability that \(x\) is at most 12 is 0.36. The probability that \(x\) is between 10 and 15, and between 12 and 17 are both 0.4. The probabilities are equal because each interval has the same width and uniform distribution assigns equal probability to equal widths.

Step by step solution

01

Drawing the Density Curve

The density curve for \(x\) would be a rectangle from 7.5 to 20 on the x-axis. In a uniform distribution, all outcomes are equally likely, thus the density curve would be a horizontal line (constant probability) in this interval.
02

Calculate Height of Density Curve

The height of the uniform density curve is given by \(1/(b-a)\) where \(a\) and \(b\) are the endpoints of the interval. Here, \(a = 7.5\) and \(b = 20\). Substituting these values gives us a height of \(1/(20 - 7.5) = 0.08.\)
03

Calculate Probability of \(x\) being at most 12

The probability that \(x\) is at most 12 (i.e., \(P(X \leq 12)\)) is calculated by finding the area under the density curve up to 12 on the x-axis. The area of this 'rectangle' is simply the height of the curve (in this case 0.08) times the width of the rectangle (which goes from 7.5 to 12). The width is thus \(12 - 7.5 = 4.5\). Thus, the required probability is \(0.08 * 4.5 = 0.36\).
04

Calculate Probability of \(x\) being between 10 and 15, and between 12 and 17

The probability that \(x\) lies between 10 and 15, and between 12 and 17 are both calculated in the same way as in step 3. The width of both intervals is 5 (15 - 10 = 5 and 17 - 12 = 5). Thus, the probabilities are both \(0.08 * 5 = 0.4\). They are equal because each interval has the same width (5 units) and the height of the density function is constant.

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Key Concepts

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

Probability Density Function
A probability density function (PDF) is essential in describing the likelihood of a continuous random variable falling within a particular range of values. When we talk about uniform distribution, this concept becomes particularly straightforward. The PDF takes the shape of a flat, horizontal line that extends evenly across the specified interval.

- **Uniform distribution** describes a scenario where every outcome in a given interval has the same probability of occurrence. This means the PDF is constant within this interval.- For the given problem, the depth of the bioturbation layer is uniformly distributed between 7.5 and 20. The PDF, in this case, is a rectangle plotted between these values on the x-axis.- To determine the height of the PDF, you use the formula: \[ ext{Height} = \frac{1}{b-a} \] where \(a\) and \(b\) mark the interval's endpoints. For our problem, substituting 7.5 for \(a\) and 20 for \(b\), we get a height of 0.08.
Continuous Probability Distribution
A continuous probability distribution needs math to accurately predict probabilities over a continuous range. Unlike discrete distributions that handle countable outcomes, continuous distributions manage large ranges often expressed in intervals.- **Uniform distribution**, specifically in continuous probability, means the probability of the occurrence is consistent throughout the given range.
- The probability that a value falls within a specified range (like between 10 and 15 in our exercise) can be determined by calculating the area under the PDF over that range.- Calculating the probability of any single point in a continuous distribution gives zero. Only ranges of values have non-zero probabilities.In the exercise, calculating probabilities involved measuring the area of shapes (rectangles) that fall within intervals. For instance, the probability that the depth of the bioturbation layer falls between 12 and 17 is found using the rectangle's width \((5)\) multiplied by the PDF's constant height \( (0.08)\). Every segment is crafted the same way, emphasizing that equal intervals carry equal weight.
Bioturbation Layer Modeling
Modeling the bioturbation layer in sediment involves understanding the layer's distribution across different depths. The uniform distribution helps highlight assumptions such as:

- Consistency in the layer's thickness over a specific range, highlighting even distribution across sediment depth. - The suspicion that outside this range, measurement variance strongly deviates from uniform distribution.
Bioturbation is a complex ecological process where organisms disturb the sediment layer. This regularity, or perceived uniformity, is an essential aspect of environmental modeling, particularly with hydrophobic pollutants, as it suggests pollutant distribution over a range where uniform behavior can be reasonably assumed. An understanding of these models informs practices such as sampling designs and remediation strategies, providing insights into how environmental factors influence sediment layers. Knowing this uniformity helps scientists predict environmental interactions, guiding vital ecological conservation efforts.

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Most popular questions from this chapter

The probability distribution of \(x,\) the number of defective tires on a randomly selected automobile checked at a certain inspection station, is given in the following table: $$ \begin{array}{lccccc} x & 0 & 1 & 2 & 3 & 4 \\ p(x) & 0.54 & 0.16 & 0.06 & 0.04 & 0.20 \end{array} $$ a. Calculate the mean value of \(x\). b. Interpret the mean value of \(x\) in the context of a long sequence of observations of number of defective tires. c. What is the probability that \(x\) exceeds its mean value? d. Calculate the standard deviation of \(x\).

A business has six customer service telephone lines. Let \(x\) denote the number of lines in use at any given time. Suppose that the probability distribution of \(x\) is as follows: $$ \begin{array}{lccccccc} x & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\ p(x) & 0.10 & 0.15 & 0.20 & 0.25 & 0.20 & 0.06 & 0.04 \end{array} $$ Write each of the following events in terms of \(x,\) and then calculate the probability of each one: a. At most three lines are in use b. Fewer than three lines are in use c. At least three lines are in use d. Between two and five lines (inclusive) are in use e. Between two and four lines (inclusive) are not in use f. At least four lines are not in use

A box contains five slips of paper, marked \(\$ 1, \$ 1, \$ 1, \$ 10,\) and \(\$ 25 .\) The winner of a contest selects two slips of paper at random and then gets the larger of the dollar amounts on the two slips. Define a random variable \(w\) by \(w=\) amount awarded. Determine the probability distribution of \(w\). (Hint: Think of the slips as numbered \(1,2,3,4,\) and \(5 .\) An outcome of the experiment will consist of two of these numbers.)

A chemical supply company currently has in stock 100 pounds of a certain chemical, which it sells to customers in 5 -pound lots. Let \(x=\) the number of lots ordered by a randomly chosen customer. The probability distribution of \(x\) is as follows: $$ \begin{array}{lcccc} x & 1 & 2 & 3 & 4 \\ p(x) & 0.2 & 0.4 & 0.3 & 0.1 \end{array} $$ a. Calculate and interpret the mean value of \(x\). b. Calculate and interpret the variance and standard deviation of \(x\).

A restaurant has four bottles of a certain wine in stock. The wine steward does not know that two of these bottles (Bottles 1 and 2) are bad. Suppose that two bottles are ordered, and the wine steward selects two of the four bottles at random. Consider the random variable \(x=\) the number of good bottles among these two. a. One possible experimental outcome is (1,2) (Bottles 1 and 2 are selected) and another is (2,4) . List all possible outcomes. b. What is the probability of each outcome in Part (a)? c. The value of \(x\) for the (1,2) outcome is 0 (neither selected bottle is good), and \(x=1\) for the outcome (2,4) . Determine the \(x\) value for each possible outcome. Then use the probabilities in Part (b) to determine the probability distribution of \(x\). (Hint: See Example 6.5 )

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