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?

Question
Modeling data distributions
asked 2020-11-20
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?

Answers (1)

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: \(\displaystyle{f{{\left({x}\right)}}}={\frac{{{1}}}{{{20}-{7.5}}}}={\frac{{{1}}}{{{12.5}}}}={0.08}\) and zero otherwise. Then f(x) can be written as: \(\displaystyle{f{{\left({x}\right)}}}{b}{e}{g}\in{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}{0.08}&{7.5}{<}{x}{<}{20}\backslash{0}&\text{otherwise}{e}{n}{d}{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}\)</span> The mean value of the given distribution can be given as: \(\displaystyle{E}{\left({X}\right)}={\int_{{-\infty}}^{{\infty}}}{x}\cdot{f{{\left({x}\right)}}}\cdot{\left.{d}{x}\right.}\) ZSK \(\displaystyle={\int_{{{7.5}}}^{{{20}}}}{x}\cdot{\left({0.08}\right)}\cdot{\left.{d}{x}\right.}\)
\(\displaystyle={0.08}{\int_{{{7.5}}}^{{{20}}}}{x}\cdot{\left.{d}{x}\right.}\)
\(\displaystyle={0.08}{{\left[{\frac{{{x}^{{{2}}}}}{{{2}}}}\right]}_{{{7.5}}}^{{{20}}}}\)
\(\displaystyle={0.08}{\left[{\frac{{{20}^{{{2}}}}}{{{2}}}}-{\frac{{{7.5}^{{{2}}}}}{{{2}}}}\right]}\)
\(\displaystyle{E}{\left({X}\right)}={13.75}\) Definition: The expected or mean value of a continuous rv X with pdf f(x) is \(\displaystyle\mu={E}{\left({X}\right)}={\int_{{-\infty}}^{{\infty}}}{x}\cdot{f{{\left({x}\right)}}}\cdot{\left.{d}{x}\right.}\) Step 2 For the pdf f(x), to calculate variance, we first calculate \(\displaystyle{E}{\left({X}^{{{2}}}\right)}\): \(\displaystyle{E}{\left({X}\right)}={\int_{{-\infty}}^{{\infty}}}{x}\cdot{f{{\left({x}\right)}}}\cdot{\left.{d}{x}\right.}\) ZSK \(\displaystyle={\int_{{{7.5}}}^{{{20}}}}{x}^{{{2}}}\cdot{\left({0.08}\right)}\cdot{\left.{d}{x}\right.}\)
\(\displaystyle={\left({0.08}\right)}{\int_{{{7.5}}}^{{{20}}}}{x}^{{{2}}}\cdot{\left.{d}{x}\right.}\)
\(\displaystyle={\left({0.08}\right)}{{\left[{\frac{{{x}^{{{3}}}}}{{{3}}}}\right]}_{{{7.5}}}^{{{20}}}}\)
\(\displaystyle={\frac{{{0.08}}}{{{3}}}}{\left[{20}^{{{3}}}-{7.5}^{{{3}}}\right]}={\frac{{{0.08}}}{{{3}}}}{\left[{8000}-{421.875}\right]}\)
\(\displaystyle{E}{\left({X}^{{{2}}}\right)}={202.08}\) As we have already calculated E(X) in the last part: \(\displaystyle{E}{\left({X}\right)}={13.75}\), hence we use following proposition: Proposition: PSKV(X)=E(X^{2})-E(X)^{2} Using this, we can write: \(\displaystyle{V}{\left({X}\right)}={202.08}-{\left({13.75}\right)}^{{{2}}}\)
\(\displaystyle{V}{\left({X}\right)}={13}\) Definition: If X is a continuous rv with pdf f(x) and h(X) is any function of X, then \(\displaystyle{E}{\left({h}{\left({x}\right)}\right)}={\int_{{-\infty}}^{{\infty}}}{h}{\left({x}\right)}\cdot{\left.{d}{x}\right.}\) 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 \(\displaystyle{F}{\left({x}\right)}={P}{\left({X}\leq{x}\right)}={\int_{{-\infty}}^{{{x}}}}{f{{\left({y}\right)}}}\cdot{\left.{d}{y}\right.}\) pdf f(x) is given to us as: \(\displaystyle{f{{\left({x}\right)}}}={b}{e}{g}\in{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}{0.08}&{7.5}{<}{x}{<}{20}\backslash{0}&{o}{t}{h}{e}{r}{w}{i}{s}{e}{e}{n}{d}{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}\)</span> For any number x between 7.5 and 20 \(\displaystyle{F}{\left({X}\right)}={\int_{{{7.5}}}^{{{x}}}}{\left({0.08}\right)}\cdot{\left.{d}{y}\right.}\)
\(\displaystyle={\left({0.08}\right)}{\int_{{{7.5}}}^{{{x}}}}{\left.{d}{y}\right.}\)
\(\displaystyle={0.08}{{\left[{y}\right]}_{{{7.5}}}^{{{x}}}}\)
\(\displaystyle={0.08}{\left[{x}-{7.5}\right]}\)
\(\displaystyle{F}{\left({X}\right)}={0.08}{x}-{0.6}\) Thus F(X) can be given as: \(\displaystyle{F}{\left({x}\right)}={b}{e}{g}\in{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}{0}&{x}{<}{7.5}\backslash{0.08}{x}-{0.6}&{7.5}\leq{x}\leq{20}\backslash{1}&{x}{>}{20}{e}{n}{d}{\left\lbrace{c}{a}{s}{e}{s}\right\rbrace}\) Step 4 c) Probability that observed depth is at most 10 is denoted by \(\displaystyle{P}{\left({X}\leq{10}\right)}\). Then using the given cdf from part(b), we can write: \(\displaystyle{\left({X}\leq{10}\right)}={F}{\left({10}\right)}\)
\(\displaystyle={0.08}{\left({10}\right)}-{0.6}\)
\(\displaystyle{P}{\left({X}\leq{10}\right)}={0.2}\) Probability that observed depth is between 10 and 15 is denoted by PSKP(10 \(\displaystyle{P}{\left({10}{<}{X}{<}{15}\right)}={F}{\left({15}\right)}-{F}{\left({10}\right)}\)</span>
\(\displaystyle={\left[{0.08}{\left({15}\right)}-{0.6}\right]}-{\left[{0.08}{\left({10}\right)}-{0.6}\right]}\)
\(\displaystyle={\left({0.6}\right)}-{\left({0.2}\right)}\)
\(\displaystyle{P}{\left({10}{<}{X}{<}{15}\right)}={0.4}\)</span> Proposition: Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any number a, \(\displaystyle{P}{\left({x}\leq{a}\right)}={F}{\left({a}\right)}\) Proposition: Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any numbers a and b with \(\displaystyle{a}{<}{b}\)</span>
\(\displaystyle{P}{\left(\leq{X}\leq{b}\right)}={F}{\left({b}\right)}-{F}{\left({a}\right)}\) Step 5 d) From part (a), we know the mean and variance of the given pdf as: \(\displaystyle{E}{\left({X}\right)}=\mu={13.75}\)
\(\displaystyle{V}{\left({X}\right)}={13}\) Now, standard deviation \(\displaystyle{\left(\sigma{x}\right)}\) can be written as: \(\displaystyle\sigma{x}=\sqrt{{{V}{\left({X}\right)}}}\)
\(\displaystyle=\sqrt{{{13}}}\)
\(\displaystyle\sigma{x}={3.6056}\) The probability the observed depth is within 1 standard deviation of the mean value is denoted by \(\displaystyle{P}{\left(\mu-\sigma{x}{<}{X}{<}\mu+\sigma{x}\right)}\)</span>
\(\displaystyle{P}{\left(\mu-\sigma{x}{<}{X}{<}\mu+\sigma{x}\right)}={P}{\left({10.6844}{<}{X}{<}{17.3556}\right)}\)</span>
\(\displaystyle={F}{\left({17.3556}\right)}-{F}{\left({10.6844}\right)}\)
\(\displaystyle={\left[{0.08}{\left({17.3556}\right)}-{0.6}\right]}-{\left[{0.08}{\left({10.6844}\right)}-{0.6}\right]}\)
\(\displaystyle={\left({0.7884}\right)}-{\left({0.2548}\right)}\)
\(\displaystyle{P}{\left(\mu-\sigma{x}{<}{X}{<}\mu+\sigma{x}\right)}={0.5336}\)</span> The probability the observed depth is within 2 standard deviations of the mean value is denoted by \(\displaystyle{P}{\left(\mu-{2}\sigma{x}{<}{X}{<}\mu+{2}\sigma{x}\right)}\)</span>
\(\displaystyle={F}{\left({20.9612}\right)}-{F}{\left({6.5388}\right)}\)
\(\displaystyle={\left[{1}\right]}-{\left[{0}\right]}\)
\(\displaystyle{P}{\left(\mu-{2}\sigma{x}{<}{X}{<}\mu+{2}\sigma{x}\right)}={1}\)</span>
0

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\(\begin{array}{|c|c|} \hline x & Mileage \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}\)
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\(\begin{array}{|c|c|} \hline x & Mileage & f(x) \\ \hline 28 & 45 \\ \hline 30 & 51\\ \hline 32 & 56\\ \hline 34 & 50\\ \hline 36 & 46\\ \hline \end{array}\)
​(Round to one decimal place as​ needed.)
\(A. 20602060xf(x)\)
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,45), (30,51), (32,56), (34,50), and (36,46). A parabola opens downward and passes through the points (28,45.7), (30,52.4), (32,54.7), (34,52.6), and (36,46.0). All points are approximate.
\(B. 20602060xf(x)\)
Acoordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2.
Data points are plotted at (43,30), (45,36), (47,41), (49,35), and (51,31). A parabola opens downward and passes through the points (43,30.7), (45,37.4), (47,39.7), (49,37.6), and (51,31). All points are approximate.
\(C. 20602060xf(x)\)
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (43,45), (45,51), (47,56), (49,50), and (51,46). A parabola opens downward and passes through the points (43,45.7), (45,52.4), (47,54.7), (49,52.6), and (51,46.0). All points are approximate.
\(D.20602060xf(x)\)
A coordinate system has a horizontal x-axis labeled from 20 to 60 in increments of 2 and a vertical y-axis labeled from 20 to 60 in increments of 2. Data points are plotted at (28,30), (30,36), (32,41), (34,35), and (36,31). A parabola opens downward and passes through the points (28,30.7), (30,37.4), (32,39.7), (34,37.6), and (36,31). All points are approximate.
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\(\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.\) and for 35
\(\displaystyle​\frac{{{l}{b}{s}}}{{{s}{q}}}\in.\)
The mileage for the tire pressure \(\displaystyle{29}\frac{{{l}{b}{s}}}{{{s}{q}}}\in.\) is
The mileage for the tire pressure \(\displaystyle{35}\frac{{{l}{b}{s}}}{{{s}{q}}}\) in. is
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