Question

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

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

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.

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 \(1/(20 - 7.5) = 0.08.\)

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 \(0.08 * 4.5 = 0.36\).

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.