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Chapter 4: Q14E (page 155)

The article 鈥淢odelling Sediment and Water Column Interactions for Hydrophobic Pollutants鈥� (Water Research, \({\rm{1984: 1169 - 1174}}\)) suggests the uniform distribution on the interval \({\rm{(7}}{\rm{.5,20)}}\)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 \({\rm{10}}\)? Between \({\rm{10}}\) and \({\rm{15}}\)? d. What is the probability that the observed depth is within \({\rm{1}}\) standard deviation of the mean value? Within \({\rm{2 }}\) standard deviations?

Short Answer

Expert verified

(a) The values are \({\rm{13}}{\rm{.75}}\) and \({\rm{13}}\).

(b) The value is \({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < 7}}{\rm{.5}}}\\{{\rm{0}}{\rm{.08x - 0}}{\rm{.6}}}&{{\rm{7}}{\rm{.5}} \le {\rm{x}} \le {\rm{20}}}\\{\rm{1}}&{{\rm{x > 20}}}\end{array}} \right.\)

(c) The values are \({\rm{0}}{\rm{.2}}\) and \({\rm{0}}{\rm{.4}}\).

(d)The values are \({\rm{0}}{\rm{.5336}}\) and \({\rm{1}}\).

Step by step solution

01

Define mean

The average of a group of numbers is simply defined as the mean. In statistics, the mean is also one of the indicators of central tendency.

02

Explanation

(a) Let \({\rm{f(x)}}\) be the uniform depth distribution on the provided interval \({\rm{(7}}{\rm{.5,20)}}\). As a result, the value of \({\rm{f(x)}}\) in this interval is:

\(\begin{array}{c}{\rm{f(x) = }}\frac{{\rm{1}}}{{{\rm{20 - 7}}{\rm{.5}}}}\\{\rm{ = }}\frac{{\rm{1}}}{{{\rm{12}}{\rm{.5}}}}\\{\rm{ = 0}}{\rm{.08}}\end{array}\)

Otherwise, it is zero.\({\rm{f(x)}}\)can then be written as:

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{0}}{\rm{.08}}}&{{\rm{7}}{\rm{.5 < x < 20}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

The following is a formula for calculating the mean value of a distribution:

\(\begin{array}{c}{\rm{E(X) = }}\int_{{\rm{ - }}\infty }^\infty {\rm{x}} {\rm{ \times f(x) \times dx}}\\{\rm{ = }}\int_{{\rm{7}}{\rm{.5}}}^{{\rm{20}}} {\rm{x}} {\rm{ \times (0}}{\rm{.08) \times dx}}\\{\rm{ = 0}}{\rm{.08}}\int_{{\rm{7}}{\rm{.5}}}^{{\rm{20}}} {\rm{x}} {\rm{ \times dx}}\\{\rm{ = 0}}{\rm{.08}}\left( {\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}} \right)_{{\rm{7}}{\rm{.5}}}^{{\rm{20}}}\\{\rm{ = 0}}{\rm{.08}}\left( {\frac{{{\rm{2}}{{\rm{0}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{\rm{7}}{\rm{.}}{{\rm{5}}^{\rm{2}}}}}{{\rm{2}}}} \right)\\{\rm{E(X) = 13}}{\rm{.75}}\end{array}\)

Definition: A continuous\({\rm{rv}}\)X with pdf\({\rm{f(x)}}\)has an expected or mean value.

\(\begin{array}{c}{\rm{\mu = E(X)}}\\{\rm{ = }}\int_{{\rm{ - }}\infty }^\infty {\rm{x}} {\rm{ \times f(x) \times dx}}\end{array}\)

03

Evaluating the values

To calculate variance for the pdf \({\rm{f(x)}}\), we first calculate \({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right)\):

\(\begin{array}{c}{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = }}\int_{{\rm{ - }}\infty }^\infty {{{\rm{x}}^{\rm{2}}}} {\rm{ \times f(x) \times dx}}\\{\rm{ = }}\int_{{\rm{7}}{\rm{.5}}}^{{\rm{20}}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times (0}}{\rm{.08) \times dx}}\\{\rm{ = (0}}{\rm{.08)}}\int_{{\rm{7}}{\rm{.5}}}^{{\rm{20}}} {{{\rm{x}}^{\rm{2}}}} {\rm{ \times dx}}\\{\rm{ = (0}}{\rm{.08)}}\left( {\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{{\rm{7}}{\rm{.5}}}^{{\rm{20}}}\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.08}}}}{{\rm{3}}}\left( {{\rm{2}}{{\rm{0}}^{\rm{3}}}{\rm{ - 7}}{\rm{.}}{{\rm{5}}^{\rm{3}}}} \right)\\{\rm{ = }}\frac{{{\rm{0}}{\rm{.08}}}}{{\rm{3}}}{\rm{(8000 - 421}}{\rm{.875)}}\\{\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 202}}{\rm{.08}}\end{array}\)

We use the following proposition because we had calculated\({\rm{E(X)}}\)in the previous section:\({\rm{E}}\left( {\rm{X}} \right){\rm{ = 13}}{\rm{.75}}\).

\({\rm{V(X) = E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ - E(X}}{{\rm{)}}^{\rm{2}}}\)

This allows us to write:

\(\begin{array}{c}{\rm{V(X) = 202}}{\rm{.08 - (13}}{\rm{.75}}{{\rm{)}}^{\rm{2}}}\\{\rm{V(X) = 13}}\end{array}\)

\({\rm{h(X)}}\)is any function of X if X is a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\).

\({\rm{E(h(x)) = }}\int_{{\rm{ - }}\infty }^\infty {\rm{h}} {\rm{(x) \times f(x) \times dx}}\)

Therefore, the values are \({\rm{13}}{\rm{.75}}\) and \({\rm{13}}\).

04

Explanation

(b) Recall the cdf of a continuous variable definition.

For each number x, the cumulative distribution function\({\rm{F(x)}}\)for a continuous \({\rm{rv}}\)\({\rm{X}}\)is defined.

\(\begin{array}{c}{\rm{F(x) = P(X}} \le {\rm{x)}}\\{\rm{ = }}\int_{{\rm{ - }}\infty }^{\rm{x}} {\rm{f}} {\rm{(y) \times dy}}\end{array}\)

We are given the following pdf\({\rm{f(x)}}\):

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{l}}{{\rm{0}}{\rm{.08}}}&{{\rm{7}}{\rm{.5 < x < 20}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

For each number x between\({\rm{7}}{\rm{.5}}\)and\({\rm{20}}\).

\(\begin{array}{c}{\rm{F(X) = }}\int_{{\rm{7}}{\rm{.5}}}^{\rm{x}} {{\rm{(0}}{\rm{.08)}}} {\rm{ \times dy}}\\{\rm{ = (0}}{\rm{.08)}}\int_{{\rm{7}}{\rm{.5}}}^{\rm{x}} {\rm{d}} {\rm{y}}\\{\rm{ = 0}}{\rm{.08(y)}}_{{\rm{7}}{\rm{.5}}}^{\rm{x}}\\{\rm{ = 0}}{\rm{.08(x - 7}}{\rm{.5)}}\\{\rm{F(X) = 0}}{\rm{.08x - 0}}{\rm{.6}}\end{array}\)

As a result,\({\rm{F(X)}}\)can be written as:

\({\rm{F(x) = }}\left\{ {\begin{array}{*{20}{l}}{\rm{0}}&{{\rm{x < 7}}{\rm{.5}}}\\{{\rm{0}}{\rm{.08x - 0}}{\rm{.6}}}&{{\rm{7}}{\rm{.5}} \le {\rm{x}} \le {\rm{20}}}\\{\rm{1}}&{{\rm{x > 20}}}\end{array}} \right.\)

05

Explanation


(c) As, \({\rm{P(X}} \le {\rm{10)}}\) denotes the probability that the depth observed is at most ten. We can then write using the cdf provided in part(b).

\(\begin{array}{c}{\rm{P(X}} \le {\rm{10) = F(10)}}\\{\rm{ = 0}}{\rm{.08(10) - 0}}{\rm{.6}}\\{\rm{P(X}} \le {\rm{10) = 0}}{\rm{.2}}\end{array}\)

\({\rm{P(10 < X < 15)}}\)denotes the probability that the observed depth is between\({\rm{10}}\)and\({\rm{15}}\).

\(\begin{array}{c}{\rm{P(10 < X < 15) = F(15) - F(10)}}\\{\rm{ = (0}}{\rm{.08(15) - 0}}{\rm{.6) - (0}}{\rm{.08(10) - 0}}{\rm{.6)}}\\{\rm{ = (0}}{\rm{.6) - (0}}{\rm{.2)}}\\{\rm{P(10 < X < 15) = 0}}{\rm{.4}}\end{array}\)

Let X be a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F}}\)as parameters\({\rm{(x)}}\). After that, for any number a,

\({\rm{P(X}} \le {\rm{a) = F(a)}}\)

Let X be a continuous\({\rm{rv}}\)with pdf\({\rm{f(x)}}\)and cdf\({\rm{F}}\)as a proposition\({\rm{(x)}}\). Then, using\({\rm{a < b}}\), for any two numbers a and b,

\({\rm{P(a}} \le {\rm{X}} \le {\rm{b) = F(b) - F(a)}}\)

Therefore, the values are \({\rm{0}}{\rm{.2}}\) and \({\rm{0}}{\rm{.4}}\).

06

Explanation


(d) We may estimate the mean and variance of the provided pdf from part(a):

\(\begin{array}{c}{\rm{E(X) = \mu }}\\{\rm{ = 13}}{\rm{.75}}\\{\rm{V(X) = 13}}\end{array}\)

Standard deviation\(\left( {{{\rm{\sigma }}_{\rm{X}}}} \right)\)can now be represented as follows:

\(\begin{array}{c}{{\rm{\sigma }}_{\rm{X}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} \\{\rm{ = }}\sqrt {{\rm{13}}} \\{{\rm{\sigma }}_{\rm{X}}}{\rm{ = 3}}{\rm{.6056}}\end{array}\)

\({\rm{P}}\left( {{\rm{\mu - }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + }}{{\rm{\sigma }}_{\rm{X}}}} \right)\)stands for the chance that the depth observed is within one standard deviation of the mean value.

\(\begin{array}{c}{\rm{P}}\left( {{\rm{\mu - }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + }}{{\rm{\sigma }}_{\rm{X}}}} \right){\rm{ = P(10}}{\rm{.6844 < X < 17}}{\rm{.3556)}}\\{\rm{ = F(17}}{\rm{.3556) - F(10}}{\rm{.6844)}}\\{\rm{ = (0}}{\rm{.08(17}}{\rm{.3556) - 0}}{\rm{.6) - (0}}{\rm{.08(10}}{\rm{.6844) - 0}}{\rm{.6)}}\\{\rm{ = (0}}{\rm{.7884) - (0}}{\rm{.2548)}}\\{\rm{P}}\left( {{\rm{\mu - }}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + }}{{\rm{\sigma }}_{\rm{X}}}} \right){\rm{ = 0}}{\rm{.5336}}\end{array}\)

\({\rm{P}}\left( {{\rm{\mu - 2}}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + 2}}{{\rm{\sigma }}_{\rm{X}}}} \right)\)denotes the probability that the observed depth is within two standard deviations of the mean value.

\(\begin{array}{c}{\rm{P}}\left( {{\rm{\mu - 2}}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + 2}}{{\rm{\sigma }}_{\rm{X}}}} \right){\rm{ = P(6}}{\rm{.5388 < X < 20}}{\rm{.9612)}}\\{\rm{ = F(20}}{\rm{.9612) - F(6}}{\rm{.5388)}}\\{\rm{ = (1) - (0)}}\\{\rm{P}}\left( {{\rm{\mu - 2}}{{\rm{\sigma }}_{\rm{X}}}{\rm{ < X < \mu + 2}}{{\rm{\sigma }}_{\rm{X}}}} \right){\rm{ = 1}}\end{array}\)

Therefore, the values are \({\rm{0}}{\rm{.5336}}\) and \({\rm{1}}\).

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Most popular questions from this chapter

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