The article “Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants” (Water Research, 1984: 1169-1174) assumes a uniform distribution over 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 average and spread of depth? b. What 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?

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svartmaleJ · Accepted Answer

Step 1a) Let f(x) be the uniform distribution of depth on the interval (7.5, 20) given to us. Therefore, the value of f(x) in this interval is equal to:f(x)=120−7.5=112.5=0.08and zero otherwise. Then f(x) can be written as:f(x){0.087.5&amp;lt;x&amp;lt;200otherwise&amp;nbsp;The average value of the given distribution can be given as:E(X)=∫−∞∞x⋅f(x)⋅dx=∫7.520x⋅(0.08)⋅dx=0.08∫7.520x⋅dx=0.08[x22]7.520=0.08[2022−7.522]E(X)=13.75Definition: The expected or mean value of a continuous rv X with pdf f(x) isμ=E(X)=∫−∞∞x⋅f(x)⋅dxStep 2For the pdf f(x), to calculate variance, we first calculate&amp;nbsp;E(X2):E(X)=∫−∞∞x⋅f(x)⋅dx&amp;nbsp;&amp;nbsp;=∫7.520x2⋅(0.08)⋅dx=(0.08)∫7.520x2⋅dx=(0.08)[x33]7.520=0.083[203−7.53]=0.083[8000−421.875]E(X2)=202.08As we have already calculated E(X) in the last part:&amp;nbsp;E(X)=13.75, hence we use following proposition: Proposition:&amp;nbsp;(V(X)=E(X2)−E(X)2Using this, we can write:&amp;nbsp;V(X)=202.08−(13.75)2V(X)=13Definition: If X is a continuous rv with pdf f(x) and h(X) is any function of X, then&amp;nbsp;E(h(x))=∫−∞∞h(x)⋅dxb) We recall the definition of a continuous variable. Definition: the cumulative distribution function F(x) for a continuous rv X is defined for every number x by&amp;nbsp;F(x)=P(X≤x)=∫−∞xf(y)⋅dy&amp;nbsp;pdf f(x) is given to us as:&amp;nbsp;f(x){0.087.5&amp;lt;x&amp;lt;200otherwise&amp;nbsp;For any number x between 7.5 and 20F(X)=∫7.5x(0.08)⋅dy=(0.08)∫7.5xdy=0.08[y]7.5x=0.08[x−7.5]F(X)=0.08x−0.6Thus F(X) can be given as:F(x)={0x&amp;lt;7.50.08x−0.67.5≤x≤201x&amp;gt;20&amp;nbsp;Step 4c) Probability that observed depth is at most 10 is denoted by&amp;nbsp;P(X≤10). Then using the given cdf from part(b), we can write:&amp;nbsp;(X≤10)=F(10)=0.08(10)−0.6P(X≤10)=0.2Probability that observed depth is between 10 and 15 is denoted by&amp;nbsp;P(10&amp;lt;X&amp;lt;15)&amp;nbsp;

The article “Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants” (Water Research, 1984: 1169-1174) suggests the uniform distri

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