The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research, $1984: 1169-1174$ ) suggests the uniform dis

Given data:
*The interval of the uniform distribution is ${\displaystyle [a,b]=[7.5,20]}$
a. The mean of the depth if the function follows a uniform distribution on the interval is,
${\displaystyle \mu =\frac{b+a}{2}}$
${\displaystyle =\frac{7.5+20}{2}}$
${\displaystyle =13.75}$ cm
The variance of the depth is given by:
${\displaystyle {\sigma }^2=\frac{1}{12}\times {(b-a)}^2}$
${\displaystyle =\frac{1}{12}\times {(20-7.5)}^2}$
${\displaystyle =13.02}$ cm
b)
The cdf or cumulative distribution function of the depth is given by:
${\displaystyle F\left(x\right)=\{\left(\begin{array}{c}0x<7.5\\ \frac{x-a}{b-a}=\frac{x-7.5}{20-7.5}7.5\le x\le 20\\ 1x\ge 20\end{array}\right)}$
c)
The probability that observed depth is at most 10 is given by:
${\displaystyle P\left((X<10)\right)}$
${\displaystyle =F\left(10\right)}$
${\displaystyle =\frac{10-7.5}{20-7.5}}$
${\displaystyle =0.2}$
The probability that observed depth is in between 10 and 15 is given by:
${\displaystyle P(10\le X\le 15)=F\left(15\right)-F\left(10\right)}$
Substitute the values in the above expression.
${\displaystyle P=F\left(15\right)-F\left(10\right)}$
${\displaystyle =\frac{15-7.5}{20-7.5}-\frac{10-7.5}{20-7.5}}$
${\displaystyle =0.4}$
Thus, the probability that observed depth is in between 10 and 15 is given by:
${\displaystyle P(10\le X\le 15)=0.4}$
d.
The limit within one deviation from the mean is,
${\displaystyle \mu -\sigma =13.75-3.608=10.14}$
${\displaystyle \mu +\sigma =13.75+3.608=17.358}$
The probability that the observed depth is within 1 standard deviation of the mean value is given by:
${\displaystyle P(10.14\le X\le 17.358)=F\left(17.358\right)-F\left(10.14\right)}$
Substitute the values in the above expression.
${\displaystyle P=F\left(18.358\right)-F\left(10.14\right)}$
${\displaystyle =\frac{17.258-7.5}{20-7.5}-\frac{10.14-7.5}{20-7.5}}$
${\displaystyle =0.789-0.211}$
${\displaystyle =0.578}$
Thus, the probability that the observed depth is within 1 standard deviation of the mean value is,
${\displaystyle P(10.14\le X\le 17.358)=0.578}$
The limit within two deviation from the mean is,
${\displaystyle \mu -2\sigma =13.75-2\times 3.608=6.534}$
${\displaystyle \mu +2\sigma =13.75+2\times 3.608=20.966}$
But the random variable is defined on the interval [7.5 and 20]
Thus,the probability that the observed depth is within 2 standard deviation of the mean value is,
${\displaystyle P(6.534\le X\le 20.966)=P(6.534\le X\le 7.5)+P(7.5\le X\le 20)+P(20\le X\le 20.966)}$
Substitute the values in the above expression.
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