/*! 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} Q76E The article suggests the lognorm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Q76E (page 182)

The article suggests the lognormal distribution as a model for \({\rm{S}}{{\rm{O}}_{\rm{2}}}\)concentration above a certain forest. Suppose the parameter values are \({\rm{\mu = 1}}{\rm{.9}}\)and \({\rm{\sigma = 0}}{\rm{.9}}\).

a. What are the mean value and standard deviation of concentration?

b. What is the probability that concentration is at most \({\rm{10}}\)? Between \({\rm{5}}\) and \({\rm{10}}\)?

Short Answer

Expert verified

(a) The mean value and standard deviation \({{\rm{\mu }}_{\rm{x}}}{\rm{ = 10}}{\rm{.02}}\)and \({{\rm{\sigma }}_{\rm{x}}}{\rm{ = 0}}{\rm{.2}}\)

(b) The probability is \(0.67\)and \(0.2955\)

Step by step solution

01

Definition of lognormal distribution

A continuous probability distribution of a random variable whose logarithm is normally distributed is known as a log-normal (or lognormal) distribution in probability theory. Y = ln (X) has a normal distribution if the random variable X is log-normally distributed.

02

Explain the mean value and standard deviation of concentration? 

Let a rv \({\rm{X}}\)indicate the \({\rm{S}}{{\rm{O}}_{\rm{2}}}\)concentration above a specific forest. Then it is assumed that \({\rm{X}}\) has the following Lognormal distribution parameters:

\(\begin{array}{l}{\rm{\mu = 1}}{\rm{.9}}\\{\rm{\sigma = 0}}{\rm{.9}}\end{array}\)

  1. We already know that the mean of a lognormal distribution is:

\(\begin{array}{c}{{\rm{\mu }}_{\rm{x}}}{\rm{ = exp}}\left( {{\rm{\mu + }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{2}}}} \right)\\{\rm{ = exp}}\left( {{\rm{1}}{\rm{.9 + }}\frac{{{{{\rm{(0}}{\rm{.9)}}}^{\rm{2}}}}}{{\rm{2}}}} \right)\\{\rm{ = exp(2}}{\rm{.305)}}\\{{\rm{\mu }}_{\rm{x}}}{\rm{ = 10}}{\rm{.02}}\end{array}\)

The Variance for lognormal distribution is given as:

\(\begin{array}{c}{\rm{V(X) = }}{{\rm{e}}^{{\rm{2\mu + }}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {{{\rm{e}}^{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ - 1}}} \right)\\{\rm{ = }}{{\rm{e}}^{{\rm{2(1}}{\rm{.9) + 0}}{\rm{.}}{{\rm{9}}^{\rm{2}}}}}{\rm{ \times }}\left( {{{\rm{e}}^{{\rm{0}}{\rm{.}}{{\rm{9}}^{\rm{2}}}}}{\rm{ - 1}}} \right)\\{\rm{ = }}{{\rm{e}}^{{\rm{4}}{\rm{.61}}}}{\rm{ \times }}\left( {{{\rm{e}}^{{\rm{0}}{\rm{.81}}}}{\rm{ - 1}}} \right)\\{\rm{V(X) = 125}}{\rm{.39}}\end{array}\)

The standard deviation for a lognormal distribution is then calculated as follows:

\(\begin{array}{c}{{\rm{\sigma }}_{\rm{x}}}{\rm{ = }}\sqrt {{\rm{V(X)}}} \\{\rm{ = }}\sqrt {{\rm{125}}{\rm{.39}}} \\{{\rm{\sigma }}_{\rm{x}}}{\rm{ = 11}}{\rm{.20}}\end{array}\)

03

Find the probability that the total number of errors?

b.\({\rm{P(X}} \le 10)\) is the probability that the concentration is at most 10. If \({\rm{X}}\) has a lognormal distribution, we know that \({\rm{ln(X)}}\) is normally distributed.

\(\begin{array}{c}{\rm{P(X}} \le 10){\rm{ = P(ln(X)}} \le \ln (10))\\{\rm{ = P(ln(X)}} \le 2.30)\\{\rm{ = P}}\left( {\frac{{\ln (X) - 1.9}}{{0.9}} \le \frac{{2.3 - 1.9}}{{0.9}}} \right)\\{\rm{ = P(Z}} \le 0.44)\\ = \phi (0.44)\end{array}\)

The cdf for the standard normal distribution rv is \(\phi {\rm{(Z)}}\). Appendix A-3 is used.

\({\rm{P(X}} \le 10) = 0.67\)

\({\rm{P(}}5 \le {\rm{X}} \le 10)\)is the probability that the concentration is between \(5\) and \(10\). Then we'll be able to write

\(\begin{array}{c}P(5 \le X \le 10){\rm{ = P(ln(5}}) \le {\rm{ln(X)}} \le \ln (10))\\{\rm{ = P(1}}{\rm{.61}} \le \ln ({\rm{X}}) \le 2.30)\\ = P\left( {\frac{{1.61 - 1.9}}{{0.9}} \le \frac{{\ln ({\rm{X}}) - 1.9}}{{0.9}} \le \frac{{2.3 - 1.9}}{{0.9}}} \right)\\{\rm{ = P(}} - 0.32 \le {\rm{Z}} \le 0.44)\\ = \phi (0.44) - \phi ( - 0.32)\end{array}\)

Here \(\phi ({\rm{Z}})\) is the cdf for standard normal distribution rv.Using Appendix A-3

\( = 0.67 - 0.3745\)

\({\rm{P(5}} \le {\rm{X}} \le 10) = 0.2955\)

Proposition : Let \({\rm{X}}\) be a continuous \({\rm{rv}}\) with \({\rm{pdff(x)}}\)and \({\rm{cdfF(x)}}\). Then for any number a,

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

And with \({\rm{a > b}}\) for any two numbers a and b.

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

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

Consider babies born in the 鈥渘ormal鈥� range of\(37 - 43\)weeks gestational age. Extensive data support the assumption that for such babies born in the United States, birth weight is normally distributed with a mean of\(3432 g\)and a standard deviation of\(482\)g. (The article 鈥淎re Babies Normal?鈥� analyzed data from a particular year; for a sensible choice of class intervals, a histogram did not look at all normal, but after further investigations it was determined that this was due to some hospitals measuring weight in grams and others measuring to the nearest ounce and then converting to grams. A modified choice of class intervals that allowed for this gave a histogram that was well described by a normal distribution.) a. What is the probability that the birth weight of a randomly selected baby of this type exceeds\(4000 g\)? Is between\(3000 and 4000 g\)? b. What is the probability that the birth weight of a randomly selected baby of this type is either less than\(2000 g\)or greater than\(5000 g\)? c. What is the probability that the birth weight of a randomly selected baby of this type exceeds\(7\)lb? d. How would you characterize the most extreme\(.1\% \)of all birth weights? e. If X is a random variable with a normal distribution and a is a numerical constant, then Y = X also has a normal distribution. Use this to determine the distribution of birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from part (c). How does this compare to your previous answer?

Mopeds (small motorcycles with an engine capacity below\({\rm{50\;c}}{{\rm{m}}^{\rm{3}}}\)) are very popular in Europe because of their mobility, ease of operation, and low cost. The article "Procedure to Verify the Maximum Speed of Automatic Transmission Mopeds in Periodic Motor Vehicle Inspections" (J. of Automobile Engr., \({\rm{2008: 1615 - 1623}}\)) described a rolling bench test for determining maximum vehicle speed. A normal distribution with mean value \({\rm{46}}{\rm{.8\;km/h}}\) and standard deviation \({\rm{1}}{\rm{.75\;km/h}}\)is postulated. Consider randomly selecting a single such moped.

a. What is the probability that maximum speed is at most\({\rm{50\;km/h}}\)?

b. What is the probability that maximum speed is at least\({\rm{48\;km/h}}\)?

c. What is the probability that maximum speed differs from the mean value by at most \({\rm{1}}{\rm{.5}}\)standard deviations?

A college professor never finishes his lecture before the end of the hour and always finishes his lectures within \({\rm{2}}\) min after the hour. Let \({\rm{X = }}\)the time that elapses between the end of the hour and the end of the lecture and suppose the pdf of \({\rm{X}}\) is

\({\rm{f(x) = \{ }}\begin{array}{*{20}{c}}{{\rm{k}}{{\rm{x}}^2}}&{{\rm{0}} \le {\rm{x}} \le {\rm{2}}}\\{\rm{0}}&{{\rm{otherwise}}}\end{array}\)

a. Find the value of \({\rm{k}}\) and draw the corresponding density curve. (Hint: Total area under the graph of \({\rm{f(x)}}\) is \({\rm{1}}\).)

b. What is the probability that the lecture ends within \({\rm{1}}\) min of the end of the hour?

c. What is the probability that the lecture continues beyond the hour for between \({\rm{60}}\) and \({\rm{90}}\) sec?

d. What is the probability that the lecture continues for at least \({\rm{90}}\) sec beyond the end of the hour?

Suppose the reaction temperature \({\rm{X}}\) (in \(^{\rm{o}}{\rm{C}}\)) in a certain chemical process has a uniform distribution with \({\rm{A = - 5}}\) and \({\rm{B = 5}}\).

a. Compute \({\rm{P(X < 0)}}\).

b. Compute \({\rm{P( - 2}}{\rm{.5 < X < 2}}{\rm{.5)}}\).

c. Compute \({\rm{P( - 2}} \le {\rm{X}} \le {\rm{3)}}\).

d. For \({\rm{k}}\) satisfying \({\rm{ - 5 < k < k + 4 < 5}}\), compute \({\rm{P(k < X < k + 4)}}\).

Example \({\rm{4}}{\rm{.5}}\) introduced the concept of time headway in traffic flow and proposed a particular distribution for \({\rm{X = }}\) the headway between two randomly selected consecutive cars (sec). Suppose that in a different traffic environment, the distribution of time headway has the form

\({\rm{f(x) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{\rm{k}}}{{{{\rm{x}}^{\rm{4}}}}}}&{{\rm{x > 1}}}\\{\rm{0}}&{{\rm{x}} \le {\rm{1}}}\end{array}} \right.\)

a. Determine the value of \({\rm{k}}\) for which \({\rm{f(x)}}\) is a legitimate pdf. b. Obtain the cumulative distribution function. c. Use the cdf from (b) to determine the probability that headway exceeds \({\rm{2}}\) sec and also the probability that headway is between \({\rm{2}}\) and \({\rm{3}}\) sec. d. Obtain the mean value of headway and the standard deviation of headway. e. What is the probability that headway is within \({\rm{1}}\) standard deviation of the mean value?
See all solutions

Recommended explanations on Math Textbooks

Decision Maths

Read Explanation

Probability and Statistics

Read Explanation

Logic and Functions

Read Explanation

Pure Maths

Read Explanation

Geometry

Read Explanation

Theoretical and Mathematical Physics

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.