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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

Suppose \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. The probability distribution of \(y\) is as follows: $$ \begin{array}{lccccc} y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & 0.01 \end{array} $$ a. Calculate and interpret \(\mu_{y}\). (Hint: See Example 6.13) b. In the long run, for what percentage of cartons is the number of broken eggs less than \(\mu_{y} ?\) Does this surprise you? c. Explain why \(\mu_{y}\) is not equal to \(\frac{0+1+2+3+4}{5}=2.0\).

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 \(x\).

The Los Angeles Times (December 13,1992 ) reported that what \(80 \%\) of airline passengers like to do most on long flights is rest or sleep. Suppose that the actual percentage is exactly \(80 \%,\) and consider randomly selecting six passengers. Then \(x=\) the number among the selected six who prefer to rest or sleep is a binomial random variable with \(n=6\) and \(p=0.8\) a. Calculate \(p(4)\), and interpret this probability. b. Calculate \(p(6),\) the probability that all six selected passengers prefer to rest or sleep. c. Calculate \(P(x \geq 4)\).

An appliance dealer sells three different models of freezers having \(13.5,15.9,\) and 19.1 cubic feet of storage space. Let \(x=\) the amount of storage space purchased by the next customer to buy a freezer. Suppose that \(x\) has the following probability distribution: $$ \begin{array}{lrrr} x & 13.5 & 15.9 & 19.1 \\ p(x) & 0.2 & 0.5 & 0.3 \end{array} $$ a. Calculate the mean and standard deviation of \(x\). (Hint: See Example 6.15\()\) b. Give an interpretation of the mean and standard deviation of \(x\) in the context of observing the outcomes of many purchases.

A breeder of show dogs is interested in the number of female puppies in a litter. If a birth is equally likely to result in a male or a female puppy, give the probability distribution of the variable \(x=\) number of female puppies in a litter of size 5 .

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