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

Modeling data distributions
ANSWERED 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. 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? 2020-11-21

Step 1

a) Let f(x) be the uniform distribution of depth on the interval (7.5, 20) given to us. Hence value of f(x) in this interval is equal to:

$$f(x)=\frac{1}{20-7.5}=\frac{1}{12.5}=0.08$$

and zero otherwise. Then f(x) can be written as:

$$f(x)\begin{cases}0.08 & 7.5<x<20\\0 & \text{otherwise}\end{cases}$$

The mean value of the given distribution can be given as:

$$E(X)=\int_{-\infty}^{\infty} x\cdot f(x)\cdot dx$$

$$=\int_{7.5}^{20} x \cdot (0.08) \cdot dx$$
$$=0.08 \int_{7.5}^{20} x \cdot dx$$
$$=0.08[\frac{x^{2}}{2}]_{7.5}^{20}$$
$$=0.08 [\frac{20^{2}}{2}-\frac{7.5^{2}}{2}]$$
$$E(X)=13.75$$

Definition: The expected or mean value of a continuous rv X with pdf f(x) is

$$\mu=E(X)=\int_{-\infty}^{\infty} x\cdot f(x)\cdot dx$$

Step 2

For the pdf f(x), to calculate variance, we first calculate $$E(X^{2})$$:

$$E(X)=\int_{-\infty}^{\infty} x\cdot f(x)\cdot dx$$

$$=\int_{7.5}^{20} x^{2} \cdot (0.08) \cdot dx$$
$$=(0.08) \int_{7.5}^{20} x^{2} \cdot dx$$
$$=(0.08)[\frac{x^{3}}{3}]_{7.5}^{20}$$
$$=\frac{0.08}{3}[20^{3}-7.5^{3}]=\frac{0.08}{3}[8000-421.875]$$
$$E(X^{2})=202.08$$

As we have already calculated E(X) in the last part: $$E(X)=13.75$$, hence we use following proposition: Proposition: $$(V(X)=E(X^{2})-E(X)^{2}$$

Using this, we can write: $$V(X)=202.08-(13.75)^{2}$$
$$V(X)=13$$

Definition: If X is a continuous rv with pdf f(x) and h(X) is any function of X, then $$E(h(x))=\int_{-\infty}^{\infty} h(x)\cdot dx$$

b) 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 $$F(x)=P(X\leq x)=\int_{-\infty}^{x} f(y)\cdot dy$$ pdf f(x) is given to us as: $$f(x)\begin{cases}0.08 & 7.5<x<20\\0 & \text{otherwise}\end{cases}$$

For any number x between 7.5 and 20

$$F(X)=\int_{7.5}^{x} (0.08)\cdot dy$$
$$=(0.08) \int_{7.5}^{x} dy$$
$$=0.08[y]_{7.5}^{x}$$
$$=0.08[x-7.5]$$
$$F(X)=0.08x-0.6$$

Thus F(X) can be given as:

$$F(x)=\begin{cases}0 & x < 7.5\\ 0.08x-0.6 & 7.5 \leq x \leq 20\\ 1 & x > 20 \end{cases}$$

Step 4

c) Probability that observed depth is at most 10 is denoted by $$P(X \leq 10)$$. Then using the given cdf from part(b), we can write: $$(X \leq 10)=F(10)$$
$$=0.08(10)-0.6$$
$$P(X \leq 10) = 0.2$$

Probability that observed depth is between 10 and 15 is denoted by $$P(10 <X < 15)$$
$$=[0.08(15)-0.6]-[0.08(10)-0.6]$$
$$=(0.6)-(0.2)$$
Proposition: Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any number a, $$P(x \leq a)= F(a)$$

Proposition: Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any numbers a and b with
$$P(\leq X \leq b)=F(b)-F(a)$$

Step 5

d) From part (a), we know the mean and variance of the given pdf as: $$E(X)=\mu=13.75$$
$$V(X)=13$$

Now, standard deviation $$(\sigma x)$$ can be written as: $$\sigma x=\sqrt{V(X)}$$
$$= \sqrt{13}$$
$$\sigma x = 3.6056$$

The probability the observed depth is within 1 standard deviation of the mean value is denoted by $$P(\mu- \sigma x < X < \mu + \sigma x)$$
$$(P(\mu - \sigma x < X<\mu+ \sigma x)=P(10.6844$$
$$=F(17.3556)-F(10.6844)$$
$$=[0.08(17.3556)-0.6]-[0.08(10.6844)-0.6]$$
$$=(0.7884)-(0.2548)$$
$$P(\mu-\sigma x< X< \mu + \sigma x)=0.5336$$

The probability the observed depth is within 2 standard deviations of the mean value is denoted by $$P(\mu-2 \sigma x< X < \mu + 2 \sigma x)$$
$$=F(20.9612)-F(6.5388)$$
$$=-$$
$$P(\mu-2 \sigma x < X < \mu + 2 \sigma x)=1$$