 # The article "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants" (Water Research, $1984: 1169-1174$ ) suggests the uniform dis Ayaana Buck 2021-01-02 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.
What are the mean and variance of depth?
b. What is the cdf of depth?
What is the probability that observed depth is at most 10? Between 10 and $15 ?$
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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Given data:
*The interval of the uniform distribution is $\left[a,b\right]=\left[7.5,20\right]$
a. The mean of the depth if the function follows a uniform distribution on the interval is,
$\mu =\frac{b+a}{2}$
$=\frac{7.5+20}{2}$
$=13.75$ cm
The variance of the depth is given by:
${\sigma }^{2}=\frac{1}{12}×{\left(b-a\right)}^{2}$
$=\frac{1}{12}×{\left(20-7.5\right)}^{2}$
$=13.02$ cm
b)
The cdf or cumulative distribution function of the depth is given by:
$F\left(x\right)=\left\{\left(\begin{array}{c}0x<7.5\\ \frac{x-a}{b-a}=\frac{x-7.5}{20-7.5}7.5\le x\le 20\\ 1x\ge 20\end{array}\right)$
c)
The probability that observed depth is at most 10 is given by:
$P\left(\left(X<10\right)\right)$
$=F\left(10\right)$
$=\frac{10-7.5}{20-7.5}$
$=0.2$
The probability that observed depth is in between 10 and 15 is given by:
$P\left(10\le X\le 15\right)=F\left(15\right)-F\left(10\right)$
Substitute the values in the above expression.
$P=F\left(15\right)-F\left(10\right)$
$=\frac{15-7.5}{20-7.5}-\frac{10-7.5}{20-7.5}$
$=0.4$
Thus, the probability that observed depth is in between 10 and 15 is given by:
$P\left(10\le X\le 15\right)=0.4$
d.
The limit within one deviation from the mean is,
$\mu -\sigma =13.75-3.608=10.14$
$\mu +\sigma =13.75+3.608=17.358$
The probability that the observed depth is within 1 standard deviation of the mean value is given by:
$P\left(10.14\le X\le 17.358\right)=F\left(17.358\right)-F\left(10.14\right)$
Substitute the values in the above expression.
$P=F\left(18.358\right)-F\left(10.14\right)$
$=\frac{17.258-7.5}{20-7.5}-\frac{10.14-7.5}{20-7.5}$
$=0.789-0.211$
$=0.578$
Thus, the probability that the observed depth is within 1 standard deviation of the mean value is,
$P\left(10.14\le X\le 17.358\right)=0.578$
The limit within two deviation from the mean is,
$\mu -2\sigma =13.75-2×3.608=6.534$
$\mu +2\sigma =13.75+2×3.608=20.966$
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,
$P\left(6.534\le X\le 20.966\right)=P\left(6.534\le X\le 7.5\right)+P\left(7.5\le X\le 20\right)+P\left(20\le X\le 20.966\right)$
Substitute the values in the above expression.