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Chapter 7: Problem 23

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 with a constant height of 0.08. The probability that \(x\) is at most 12 is 0.36, and the probability that \(x\) falls between the intervals 10 and 15 and 12 and 17 is both 0.4 due to the property of uniform distribution.

Step by step solution

01

A. Drawing the density curve

Since \(x\) follows the uniform distribution on the interval \(7.5\) to \(20\), the density curve will be a rectangle with the base representing the interval \([7.5, 20]\) and the height corresponding to the constant probability density function \(f(x)\) over this interval.
02

B. Calculate the height of the density curve

The constant height of the density curve for a uniform distribution can be calculated by taking the reciprocal of the length of the interval. So for \(x\), the height \(h\) would be calculated as \(h = 1/(20-7.5) = 0.08\).
03

C. Calculate the probability that \(x\) is at most 12

For a uniform distribution, the probability that \(x\) is at most any value \(a\) within the interval can be calculated by finding the length of the interval from the lower bound to \(a\) and multiplying it by the constant height. Thus, the probability \(P(X \leq 12)\) is found by calculating \((12-7.5) * 0.08 = 0.36\).
04

D. Calculate the probability that \(x\) is between other intervals and explain equality

The probability that \(x\) is within any given interval can be calculated in the same fashion as before: by finding the length of the interval and multiplying it by the height of the density curve. Thus, \(P(10 \leq X \leq 15) = (15-10) * 0.08 = 0.4\) and \(P(12 \leq X \leq 17) = (17-12) * 0.08 = 0.4\). These two probabilities are equal because the length of the intervals is the same (5 units each) and the height of the density curve is constant. This demonstrates a property of the uniform distribution: equal-interval lengths have equal probabilities.

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