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>

\(\displaystyle={0.08}{\int_{{{7.5}}}^{{{20}}}}{x}\cdot{\left.{d}{x}\right.}\)

\(\displaystyle={0.08}{{\left[{\frac{{{x}^{{{2}}}}}{{{2}}}}\right]}_{{{7.5}}}^{{{20}}}}\)

\(\displaystyle={0.08}{\left[{\frac{{{20}^{{{2}}}}}{{{2}}}}-{\frac{{{7.5}^{{{2}}}}}{{{2}}}}\right]}\)

\(\displaystyle{E}{\left({X}\right)}={13.75}\) 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>