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:

\(f(x)=\frac{1}{20-7.5}=\frac{1}{12.5}=0.08\)

and zero otherwise. Then f(x) can be written as:

\(f(x)\begin{cases}0.08 & 7.5<x<20\\0 & \text{otherwise}\end{cases}\)

The mean value of the given distribution can be given as:

\(E(X)=\int_{-\infty}^{\infty} x\cdot f(x)\cdot dx\)

\(=\int_{7.5}^{20} x \cdot (0.08) \cdot dx\)

\(=0.08 \int_{7.5}^{20} x \cdot dx\)

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

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

\(E(X)=13.75\)

Definition: The expected or mean value of a continuous rv X with pdf f(x) is

\(\mu=E(X)=\int_{-\infty}^{\infty} x\cdot f(x)\cdot dx\)

Step 2

For the pdf f(x), to calculate variance, we first calculate \(E(X^{2})\):

\(E(X)=\int_{-\infty}^{\infty} x\cdot f(x)\cdot dx\)

\(=\int_{7.5}^{20} x^{2} \cdot (0.08) \cdot dx\)

\(=(0.08) \int_{7.5}^{20} x^{2} \cdot dx\)

\(=(0.08)[\frac{x^{3}}{3}]_{7.5}^{20}\)

\(=\frac{0.08}{3}[20^{3}-7.5^{3}]=\frac{0.08}{3}[8000-421.875]\)

\(E(X^{2})=202.08\)

As we have already calculated E(X) in the last part: \(E(X)=13.75\), hence we use following proposition: Proposition: \((V(X)=E(X^{2})-E(X)^{2}\)

Using this, we can write: \(V(X)=202.08-(13.75)^{2}\)

\(V(X)=13\)

Definition: If X is a continuous rv with pdf f(x) and h(X) is any function of X, then \(E(h(x))=\int_{-\infty}^{\infty} h(x)\cdot dx\)

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 \(F(x)=P(X\leq x)=\int_{-\infty}^{x} f(y)\cdot dy\) pdf f(x) is given to us as: \(f(x)\begin{cases}0.08 & 7.5<x<20\\0 & \text{otherwise}\end{cases}\)

For any number x between 7.5 and 20

\(F(X)=\int_{7.5}^{x} (0.08)\cdot dy\)

\(=(0.08) \int_{7.5}^{x} dy\)

\(=0.08[y]_{7.5}^{x}\)

\(=0.08[x-7.5]\)

\(F(X)=0.08x-0.6\)

Thus F(X) can be given as:

\(F(x)=\begin{cases}0 & x < 7.5\\ 0.08x-0.6 & 7.5 \leq x \leq 20\\ 1 & x > 20 \end{cases}\)

Step 4

c) Probability that observed depth is at most 10 is denoted by \(P(X \leq 10)\). Then using the given cdf from part(b), we can write: \((X \leq 10)=F(10)\)

\(=0.08(10)-0.6\)

\(P(X \leq 10) = 0.2\)

Probability that observed depth is between 10 and 15 is denoted by \(P(10 <X < 15)\)

\(=[0.08(15)-0.6]-[0.08(10)-0.6]\)

\(=(0.6)-(0.2)\)

Proposition: Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any number a, \(P(x \leq a)= F(a)\)

Proposition: Let X be a continuous rv with pdf f(x) and cdf F(x). Then for any numbers a and b with

\(P(\leq X \leq b)=F(b)-F(a)\)

Step 5

d) From part (a), we know the mean and variance of the given pdf as: \(E(X)=\mu=13.75\)

\(V(X)=13\)

Now, standard deviation \((\sigma x)\) can be written as: \(\sigma x=\sqrt{V(X)}\)

\(= \sqrt{13}\)

\(\sigma x = 3.6056\)

The probability the observed depth is within 1 standard deviation of the mean value is denoted by \(P(\mu- \sigma x < X < \mu + \sigma x)\)

\((P(\mu - \sigma x < X<\mu+ \sigma x)=P(10.6844\)

\(=F(17.3556)-F(10.6844)\)

\(=[0.08(17.3556)-0.6]-[0.08(10.6844)-0.6]\)

\(=(0.7884)-(0.2548)\)

\(P(\mu-\sigma x< X< \mu + \sigma x)=0.5336\)

The probability the observed depth is within 2 standard deviations of the mean value is denoted by \(P(\mu-2 \sigma x< X < \mu + 2 \sigma x)\)

\(=F(20.9612)-F(6.5388)\)

\(=[1]-[0]\)

\(P(\mu-2 \sigma x < X < \mu + 2 \sigma x)=1\)