Given data:

*The interval of the uniform distribution is \(\displaystyle{\left[{a},{b}\right]}={\left[{7.5},{20}\right]}\)

a. The mean of the depth if the function follows a uniform distribution on the interval is,

\(\displaystyle\mu=\frac{{{b}+{a}}}{{2}}\)

\(\displaystyle=\frac{{{7.5}+{20}}}{{2}}\)

\(\displaystyle={13.75}\) cm

The variance of the depth is given by:

\(\displaystyle\sigma^{2}=\frac{1}{{12}}\times{\left({b}-{a}\right)}^{2}\)

\(\displaystyle=\frac{1}{{12}}\times{\left({20}-{7.5}\right)}^{2}\)

\(\displaystyle={13.02}\) cm

b)

The cdf or cumulative distribution function of the depth is given by:

\(\displaystyle{F}{\left({x}\right)}={\left\lbrace{\left(\begin{matrix}{0}{x}<{7.5}\\\frac{{{x}-{a}}}{{{b}-{a}}}=\frac{{{x}-{7.5}}}{{{20}-{7.5}}}{7.5}\le{x}\le{20}\\{1}{x}\ge{20}\end{matrix}\right)}\right.}\)</span>

c)

The probability that observed depth is at most 10 is given by:

\(\displaystyle{P}{\left({\left({X}<{10}\right)}\right)}\)</span>

\(\displaystyle={F}{\left({10}\right)}\)

\(\displaystyle=\frac{{{10}-{7.5}}}{{{20}-{7.5}}}\)

\(\displaystyle={0.2}\)

The probability that observed depth is in between 10 and 15 is given by:

\(\displaystyle{P}{\left({10}\le{X}\le{15}\right)}={F}{\left({15}\right)}-{F}{\left({10}\right)}\)

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:

\(\displaystyle{P}{\left({10}\le{X}\le{15}\right)}={0.4}\)

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:

\(\displaystyle{P}{\left({10.14}\le{X}\le{17.358}\right)}={F}{\left({17.358}\right)}-{F}{\left({10.14}\right)}\)

Substitute the values in the above expression.

\(\displaystyle{P}={F}{\left({18.358}\right)}-{F}{\left({10.14}\right)}\)

\(\displaystyle=\frac{{{17.258}-{7.5}}}{{{20}-{7.5}}}-\frac{{{10.14}-{7.5}}}{{{20}-{7.5}}}\)

\(\displaystyle={0.789}-{0.211}\)

\(\displaystyle={0.578}\)

Thus, the probability that the observed depth is within 1 standard deviation of the mean value is,

\(\displaystyle{P}{\left({10.14}\le{X}\le{17.358}\right)}={0.578}\)

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,

\(\displaystyle{P}{\left({6.534}\le{X}\le{20.966}\right)}={P}{\left({6.534}\le{X}\le{7.5}\right)}+{P}{\left({7.5}\le{X}\le{20}\right)}+{P}{\left({20}\le{X}\le{20.966}\right)}\)

Substitute the values in the above expression.

\(\displaystyle{P}{\left({6.534}\le{X}\le{20.966}\right)}={0}+{1}+{0}={1}\)

Thus, the probability that the observed depth is within 2 standard deviation of the mean value is,

\(\displaystyle{P}{\left({6.534}\le{X}\le{20.966}\right)}={1}\)

*The interval of the uniform distribution is \(\displaystyle{\left[{a},{b}\right]}={\left[{7.5},{20}\right]}\)

a. The mean of the depth if the function follows a uniform distribution on the interval is,

\(\displaystyle\mu=\frac{{{b}+{a}}}{{2}}\)

\(\displaystyle=\frac{{{7.5}+{20}}}{{2}}\)

\(\displaystyle={13.75}\) cm

The variance of the depth is given by:

\(\displaystyle\sigma^{2}=\frac{1}{{12}}\times{\left({b}-{a}\right)}^{2}\)

\(\displaystyle=\frac{1}{{12}}\times{\left({20}-{7.5}\right)}^{2}\)

\(\displaystyle={13.02}\) cm

b)

The cdf or cumulative distribution function of the depth is given by:

\(\displaystyle{F}{\left({x}\right)}={\left\lbrace{\left(\begin{matrix}{0}{x}<{7.5}\\\frac{{{x}-{a}}}{{{b}-{a}}}=\frac{{{x}-{7.5}}}{{{20}-{7.5}}}{7.5}\le{x}\le{20}\\{1}{x}\ge{20}\end{matrix}\right)}\right.}\)</span>

c)

The probability that observed depth is at most 10 is given by:

\(\displaystyle{P}{\left({\left({X}<{10}\right)}\right)}\)</span>

\(\displaystyle={F}{\left({10}\right)}\)

\(\displaystyle=\frac{{{10}-{7.5}}}{{{20}-{7.5}}}\)

\(\displaystyle={0.2}\)

The probability that observed depth is in between 10 and 15 is given by:

\(\displaystyle{P}{\left({10}\le{X}\le{15}\right)}={F}{\left({15}\right)}-{F}{\left({10}\right)}\)

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:

\(\displaystyle{P}{\left({10}\le{X}\le{15}\right)}={0.4}\)

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:

\(\displaystyle{P}{\left({10.14}\le{X}\le{17.358}\right)}={F}{\left({17.358}\right)}-{F}{\left({10.14}\right)}\)

Substitute the values in the above expression.

\(\displaystyle{P}={F}{\left({18.358}\right)}-{F}{\left({10.14}\right)}\)

\(\displaystyle=\frac{{{17.258}-{7.5}}}{{{20}-{7.5}}}-\frac{{{10.14}-{7.5}}}{{{20}-{7.5}}}\)

\(\displaystyle={0.789}-{0.211}\)

\(\displaystyle={0.578}\)

Thus, the probability that the observed depth is within 1 standard deviation of the mean value is,

\(\displaystyle{P}{\left({10.14}\le{X}\le{17.358}\right)}={0.578}\)

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,

\(\displaystyle{P}{\left({6.534}\le{X}\le{20.966}\right)}={P}{\left({6.534}\le{X}\le{7.5}\right)}+{P}{\left({7.5}\le{X}\le{20}\right)}+{P}{\left({20}\le{X}\le{20.966}\right)}\)

Substitute the values in the above expression.

\(\displaystyle{P}{\left({6.534}\le{X}\le{20.966}\right)}={0}+{1}+{0}={1}\)

Thus, the probability that the observed depth is within 2 standard deviation of the mean value is,

\(\displaystyle{P}{\left({6.534}\le{X}\le{20.966}\right)}={1}\)