/*! 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} Problem 11 The article "The Statistics of P... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "The Statistics of Phytotoxic Air Pollutants" (J. of Royal Stat. Soc., 1989: 183-198) suggests the lognormal distribution as a model for \(\mathrm{SO}_{2}\) concentration above a certain forest. Suppose the parameter values are \(\mu=1.9\) and \(\sigma=.9\). a. What are the mean value and standard deviation of concentration? b. What is the probability that concentration is at most 10 ? Between 5 and 10 ?

Short Answer

Expert verified
a. Mean: 10.03; SD: 11.17. b. At most 10: 0.672; Between 5 and 10: 0.299.

Step by step solution

01

Identify Parameters for Lognormal Distribution

The lognormal distribution in question is defined by its parameters \(\mu = 1.9\) and \(\sigma = 0.9\). Here, \(\mu\) and \(\sigma\) are the mean and standard deviation of the natural logarithm of the \(\mathrm{SO}_2\) concentrations, respectively.
02

Calculate the Mean of the Lognormal Distribution

The mean of a lognormal distribution is given by the formula: \( E(X) = e^{\mu + \frac{\sigma^2}{2}} \).Substitute the values, \( E(X) = e^{1.9 + \frac{0.9^2}{2}} = e^{1.9 + 0.405} = e^{2.305} \). Using a calculator, \( e^{2.305} \approx 10.03 \).
03

Calculate Standard Deviation of the Lognormal Distribution

The standard deviation of a lognormal distribution is calculated as:\( \text{SD}(X) = \sqrt{(e^{\sigma^2} - 1) \cdot e^{2\mu + \sigma^2}} \).Substituting the given values: \( \text{SD}(X) = \sqrt{(e^{0.9^2} - 1) \cdot e^{2\cdot1.9 + 0.9^2}} \).Simplifying, \( \text{SD}(X) = \sqrt{(e^{0.81} - 1) \cdot e^{4.605}} \).Calculate separately: \( e^{0.81} \approx 2.24791 \) and \( (e^{0.81} - 1) = 1.24791 \), and \( e^{4.605} \approx 100.064 \).Thus, \( \text{SD}(X) \approx \sqrt{1.24791 \times 100.064} \approx \sqrt{124.81804} \approx 11.17 \).
04

Calculate Probability That Concentration is at Most 10

For a lognormal distribution, the probability \( P(X \leq x) \) is given by \( P(\ln(X) \leq \ln(x)) \), equivalent to a normal distribution.Calculate \( Z = \frac{\ln(10) - \mu}{\sigma} = \frac{2.302 - 1.9}{0.9} \approx 0.447 \).Using the standard normal distribution table, \( P(Z \leq 0.447) \approx 0.672 \).
05

Calculate Probability That Concentration is Between 5 and 10

Calculate probabilities for each limit and subtract: - \( P(X \leq 10) \) was previously found as \( 0.672 \).- Calculate \( P(X \leq 5) \): \( Z = \frac{\ln(5) - 1.9}{0.9} = \frac{1.609 - 1.9}{0.9} \approx -0.324 \).- Using the standard normal distribution table, \( P(Z \leq -0.324) \approx 0.373 \).- Thus, \( P(5 < X \leq 10) = 0.672 - 0.373 = 0.299 \).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean and Standard Deviation
The mean and standard deviation of a lognormal distribution are unique in their calculation and differ from those of a normal distribution. Given the parameters \( \mu = 1.9 \) and \( \sigma = 0.9 \), these values describe the distribution of the natural logarithm of the data.
To find the true mean (expected value) of the distribution, we use the formula:
  • \( E(X) = e^{\mu + \frac{\sigma^2}{2}} \)
Substituting the values, \( E(X) = e^{1.9 + \frac{0.9^2}{2}} = e^{2.305} \). Calculating, we find \( E(X) \approx 10.03 \). This tells us that the average concentration of \( \mathrm{SO}_2 \) is around 10.03.
The standard deviation of a lognormal distribution is calculated using:
  • \( \text{SD}(X) = \sqrt{(e^{\sigma^2} - 1) \cdot e^{2\mu + \sigma^2}} \)
For the provided parameters, this becomes \( \text{SD}(X) \approx 11.17 \). The standard deviation indicates the level of variability or dispersion of the concentration values.
Probability Calculations
In working with a lognormal distribution, probabilities involve converting data into the logarithmic scale and then applying the normal distribution.
First, to find the probability that the \( \mathrm{SO}_2 \) concentration is at most 10, convert 10 into its logarithmic value. Compute:
  • \( Z = \frac{\ln(10) - \mu}{\sigma} = \frac{2.302 - 1.9}{0.9} \approx 0.447 \)
Using the standard normal distribution, \( P(Z \leq 0.447) \approx 0.672 \), so there's about a 67.2% chance that the concentration is less than or equal to 10.
For concentrations ranging from 5 to 10, calculate the probability for 5, which involves:
  • \( Z = \frac{\ln(5) - 1.9}{0.9} = \frac{1.609 - 1.9}{0.9} \approx -0.324 \)
The probability \( P(Z \leq -0.324) \approx 0.373 \). The probability of being between 5 and 10 is \( 0.672 - 0.373 = 0.299 \), or 29.9%.
Parameter Identification
Identifying parameters \( \mu \) and \( \sigma \) is crucial for utilizing a lognormal distribution effectively. These parameters are not measuring the raw data directly but instead the logarithm of the data.
\( \mu \) represents the mean of the log-transformed values; it indicates where the center of the log-transformed distribution is located. A value of \( \mu = 1.9 \) suggests a central position for the logarithmic transformation of \( \mathrm{SO}_2 \) concentrations.
\( \sigma \) signifies the standard deviation of the natural log transformation. At \( \sigma = 0.9 \), it reflects the spread or dispersion of these log-transformed concentrations.
Together, these parameters help convert the real-world measurements into a format where statistical methods for normal distributions apply, a necessary step before performing any probability calculations.
Lognormal Probability
Understanding Lognormal Probability is essential for interpreting data that is positively skewed, where values do not go negative.
The lognormal probability distribution applies when the natural logarithm of a dataset results in a normal distribution, which is common in various natural phenomena.
  • The concentration of \( \mathrm{SO}_2 \) above the forest fits this type, as suggested in the study referenced.
  • Knowing that it is lognormal allows us to apply techniques and formulas from normal distributions, but only after logging the data.
Given the proper context, lognormal probability provides a meaningful way to model and predict real-world phenomena that are not symmetrical and are bound on one side.
This approach is practical when dealing with chemical concentrations, time-to-failure data, and financial returns, where typical values skew to the right.

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

Suppose the time spent by a randomly selected student who uses a terminal connected to a local time-sharing computer facility has a gamma distribution with mean \(20 \mathrm{~min}\) and variance \(80 \mathrm{~min}^{2}\). a. What are the values of \(\alpha\) and \(\beta\) ? b. What is the probability that a student uses the terminal for at most 24 min? c. What is the probability that a student spends between 20 and \(40 \mathrm{~min}\) using the terminal?

The accompanying observations are precipitation values during March over a 30 -year period in Minneapolis-St. Paul. \(\begin{array}{rrrrrr}.77 & 1.20 & 3.00 & 1.62 & 2.81 & 2.48 \\ 1.74 & .47 & 3.09 & 1.31 & 1.87 & .96 \\ .81 & 1.43 & 1.51 & .32 & 1.18 & 1.89 \\ 1.20 & 3.37 & 2.10 & .59 & 1.35 & .90 \\ 1.95 & 2.20 & .52 & .81 & 4.75 & 2.05\end{array}\) a. Construct and interpret a normal probability plot for this data set. b. Calculate the square root of each value and then construct a normal probability plot based on this transformed data. Does it seem plausible that the square root of precipitation is normally distributed? c. Repeat part (b) after transforming by cube roots.

Let \(U\) have a uniform distribution on the interval \([0,1]\). Then observed values having this distribution can be obtained from a computer's random number generator. Let \(X=-(1 / \lambda) \ln (1-U)\) a. Show that \(X\) has an exponential distribution with parameter \(\lambda\). [Hint: The cdf of \(X\) is \(F(x)=P(X \leq x) ; X \leq x\) is equivalent to \(U \leq\) ? b. How would you use part (a) and a random number generator to obtain observed values from an exponential distribution with parameter \(\lambda=10\) ?

The article "Three Sisters Give Birth on the Same Day" (Chance, Spring 2001, 23-25) used the fact that three Utah sisters had all given birth on March 11, 1998 as a basis for posing some interesting questions regarding birth coincidences. a. Disregarding leap year and assuming that the other 365 days are equally likely, what is the probability that three randomly selected births all occur on March 11 ? Be sure to indicate what, if any, extra assumptions you are making. b. With the assumptions used in part (a), what is the probability that three randomly selected births all occur on the same day? c. The author suggested that, based on extensive data, the length of gestation (time between conception and birth) could be modeled as having a normal distribution with mean value 280 days and standard deviation \(19.88\) days. The due dates for the three Utah sisters were March 15 , April 1, and April 4, respectively. Assuming that all three due dates are at the mean of the distribution, what is the probability that all births occurred on March 11 ? [Hint: The deviation of birth date from due date is normally distributed with mean 0 .] d. Explain how you would use the information in part (c) to calculate the probability of a common birth date.

Let \(X\) have a binomial distribution with parameters \(n=25\) and \(p\). Calculate each of the following probabilities using the normal approximation (with the continuity correction) for the cases \(p=.5, .6\), and \(.8\) and compare to the exact probabilities calculated from Appendix Table A.1. a. \(P(15 \leq X \leq 20)\) b. \(P(X \leq 15)\) c. \(P(20 \leq X)\)

See all solutions

Recommended explanations on Math Textbooks

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.