/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q14E The article 鈥淢odelling Sedimen... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

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 \({\rm{1}}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Let X have a uniform distribution on the interval \({\rm{(A,B)}}\). a. Obtain an expression for the \({\rm{(100p)th}}\) percentile. b. Compute \({\rm{E(X),V(X)}}\) and \({{\rm{\sigma }}_{\rm{X}}}\). c. For n, a positive integer, compute \({\rm{E}}\left( {{{\rm{X}}^{\rm{n}}}} \right)\).

Let \({\rm{X}}\) denote the temperature at which a certain chemical reaction takes place. Suppose that \({\rm{X}}\) has pdf

\(f(x) = \left\{ {\begin{aligned}{{}{}}{\frac{1}{9}\left( {4 - {x^2}} \right)}&{ - 1拢x拢2} \\0&{{\text{ }}otherwise{\text{ }}}\end{aligned}} \right.\)

a. Sketch the graph of \({\rm{f(x)}}\).

b. Determine the cdf and sketch it.

c. Is \({\rm{0}}\) the median temperature at which the reaction takes place? If not, is the median temperature smaller or larger than\({\rm{0}}\)?

d. Suppose this reaction is independently carried out once in each of ten different labs and that the pdf of reaction time in each lab is as given. Let \({\rm{Y = }}\) the number among the ten labs at which the temperature exceeds \({\rm{1}}\). What kind of distribution does \({\rm{Y}}\) have? (Give the names and values of any parameters.)

An oocyte is a female germ cell involved in reproduction. Based on analyses of a large sample, the article "Reproductive Traits of Pioneer Gastropod Species Colonizing Deep-Sea Hydrothermal Vents After an Eruption" (Marine Biology, \({\rm{2011: 181 - 192}}\)) proposed the following mixture of normal distributions as a model for the distribution of \({\rm{X = }}\) oocyte diameter\({\rm{(\mu m)}}\):

\({\rm{f(x) = p}}{{\rm{f}}_{\rm{1}}}\left( {{\rm{x,}}{{\rm{\mu }}_{\rm{1}}}{\rm{,\sigma }}} \right){\rm{ + (1 - p)}}{{\rm{f}}_{\rm{2}}}\left( {{\rm{x,}}{{\rm{\mu }}_{\rm{2}}}{\rm{,\sigma }}} \right)\)

where \({{\rm{f}}_{\rm{1}}}\) and \({{\rm{f}}_{\rm{2}}}\) are normal pdfs. Suggested parameter values were\({\rm{p = }}{\rm{.35,}}{{\rm{\mu }}_{\rm{1}}}{\rm{ = 4}}{\rm{.4,}}{{\rm{\mu }}_{\rm{2}}}{\rm{ = 5}}{\rm{.0}}\), and\({\rm{\sigma = }}{\rm{.27}}\).

a. What is the expected (i.e. mean) value of oocyte diameter?

b. What is the probability that oocyte diameter is between \({\rm{4}}{\rm{.4\mu m}}\) and \({\rm{5}}{\rm{.0\mu m}}\) ? (Hint: Write an expression for the corresponding integral, carry the integral operation through to the two components, and then use the fact that each component is a normal pdf.)

c. What is the probability that oocyte diameter is smaller than its mean value? What does this imply about the shape of the density curve?

Let X denote the distance \({\rm{(m)}}\) that an animal moves from its birth site to the first territorial vacancy it encounters. Suppose that for banner-tailed kangaroo rats, X has an exponential distribution with parameter \({\rm{\lambda = }}{\rm{.01386}}\) (as suggested in the article "Competition and Dispersal from Multiple Nests," Ecology, 1997: 873-883).

a. What is the probability that the distance is at most \({\rm{100\;m}}\)? At most \({\rm{200\;m}}\) ? Between 100 and\({\rm{200\;m}}\)?

b. What is the probability that distance exceeds the mean distance by more than 2 standard deviations?

c. What is the value of the median distance?

A consumer is trying to decide between two long-distance calling plans. The first one charges a flat rate of \({\rm{10}}\) per minute, whereas the second charges a flat rate of \({\rm{99}}\) for calls up to \({\rm{20}}\) minutes in duration and then \({\rm{10\% }}\)for each additional minute exceeding \({\rm{20}}\)(assume that calls lasting a non-integer number of minutes are charged proportionately to a whole-minute's charge). Suppose the consumer's distribution of call duration is exponential with parameter\({\rm{\lambda }}\).

a. Explain intuitively how the choice of calling plan should depend on what the expected call duration is.

b. Which plan is better if expected call duration is \({\rm{10}}\) minutes? \({\rm{15}}\)minutes? (Hint: Let \({{\rm{h}}_{\rm{1}}}{\rm{(x)}}\) denote the cost for the first plan when call duration is \({\rm{x}}\) minutes and let \({{\rm{h}}_{\rm{2}}}{\rm{(x)}}\)be the cost function for the second plan. Give expressions for these two cost functions, and then determine the expected cost for each plan.)
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Geometry

Read Explanation

Logic and Functions

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.