The article “Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants” (?Water Research, ?1984: 1169–1174) suggests the uniform distribution on the interval (7.5, 20) as a model for depth (cm) of the bioturbation layer in sediment in a certain region. a.? ?What are the mean and variance of depth? b.? hat is the cdf of depth? c.? ?What is the probability that observed depth is at most 10? Between 10 and 15? d.? ?What is the probability that the observed depth is within 1 standard deviation of the mean value? Within 2 standard deviations?

Answer: Step1: A uniform distribution on the interval [A,B] if the pdf of ‘X’ is Given, the uniform distribution on the interval (7.5, 20) as a model for depth (cm) of the bioturbation layer in sediment in a certain region. That is, Step2: a). To find the mean and variance of depth. 20 Mean = E(x) = xf(x) dx 7.5 20 1 = x 207.5dx 7.5 20 = x dx 7.5 12.5 20 = 1 x dx 12.57.5 1 x1+1 20 = 12.5 1+1] 7.5 = 1 ( ) 20 12.5 2 7.5 1 2 2 = 25 ((7.5) (20) ) = 13.75. 20 V(x) = =2 (x 13.75) f(x) dx 7.5 20 = (x 13.75) 2 1 dx 207.5 7.5 2 2 2 Here we using the formula (a b) = a 2ab + b 20 = 1 (x 2(13.75) x + (13.75) ) dx2 12.5 7.5 1 x3 20 2 20 2 20 = 12.5([ 3 7.5 27.5[x ] 7.5+ (13.75) [x] 7.5 1 (20) (7.5) 2 2 = 12.5 3 ] 27.5[(20) (7.5) ] + 189.0625[20 7.5] = 1 (162.76) 12.5 = 13. Standard deviation = 13 = 3.6056. Therefore, the mean and variance of depth is E(x) = 13.75 and variance V(x) = 13. Step3: b). The aim is to obtain the cumulative distribution function. The cumulative distribution function is x F(x) = 1 dx 7.5 207.5 = 1 [x] x 12.5 7.5 1 = 12.5[x 7.5] = x7.5 12.5 20 1 F(x) = 207.5x 7.5 = 1 [x] 20 12.5 7.5 1 = 12.5[20 7.5] 12.5 = 12.5 = 1 Then, the cumulative distribution function of depth is given by Step4: c). To find the probability that observed depth is at most 10. That is, 107.5 P(x < 10) = F(10) = 12.5 = 0.2. Therefore, the probability that observed depth is at most 10 is 0.2. Then, To find the probability that observed depth is between 10 and 15. P(10 < x < 15) = F(15) - F(10) 157.5 107.5 = 12.5 - 12.5 = 0.6 - 0.2 = 0.4. Therefore, the probability that observed depth is between 10 and 15 is 0.4. Step5: d). i).To find the probability that the observed depth is within 1 standard deviation. That is, E(x) ± 1 13.75 ± 1(3.6056) (10.1444,17.3565) Therefore, P(- < X< + ) F(17.3565) - F(10.1444) 17.35657.510.14447.5 12.5 - 12.5 0.7885 0.2116 0.5769 Therefore, the probability that the observed depth is within 1 standard deviation is 0.5769 ii). Next to find the probability that the observed depth is within 2 standard deviation . That is, E(x) ± 2 13.75 ± 2(3.6056) (6.5388,20.9612). Therefore, P(-2 < X< +2 ) F(20.9612) - F(6.5388) 20.96127-56.53887.5 12.5 12.5 1.0769- (-0.0768) 1.1537 Therefore, the probability that the observed depth is within 2 standard deviation is 1.1537.