/*! 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 22 The article " Determination of t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article " Determination of the MTF of Positive Photoresists Using the Monte Carlo Method" (Photographic Sct. and Engr., 1983: 254-260) proposes the exponential distribution with parameter \(\lambda=.93\) as a model for the distribution of a photon's free path length \((\mu \mathrm{m})\) under certain circumstances. Suppose this is the correct model. a. What is the expected path length, and what is the standard deviation of path length? b. What is the probability that path length exceeds \(3.0\) ? What is the probability that path length is between \(1.0\) and \(3.0\) ? c. What value is exceeded by only \(10 \%\) of all path lengths?

Short Answer

Expert verified
a. Expected path length and standard deviation are both approximately 1.075 µm. b. Probability exceeds 3.0 is 0.0672; between 1.0 and 3.0 is 0.255. c. The value exceeded by 10% of path lengths is about 2.481 µm.

Step by step solution

01

Understanding the Problem

We need to determine certain characteristics of an exponential distribution given the parameter \( \lambda = 0.93 \). This involves finding the expected value, standard deviation, and certain probabilities related to the path length of a photon.
02

Calculate Expected Value

For an exponential distribution with parameter \( \lambda \), the expected value is given by \( E(X) = \frac{1}{\lambda} \). Plugging in \( \lambda = 0.93 \), the expected value is \( E(X) = \frac{1}{0.93} \approx 1.075 \mu m \).
03

Calculate Standard Deviation

The standard deviation for an exponential distribution is also \( \sigma = \frac{1}{\lambda} \). So the standard deviation of the path length is \( \sigma = \frac{1}{0.93} \approx 1.075 \mu m \).
04

Calculate Probability of Exceeding 3.0

The probability that a path length exceeds \(3.0\) is \( P(X > 3) = 1 - P(X \leq 3) \). For an exponential distribution, this is \( e^{-\lambda \times 3.0} = e^{-0.93 \times 3} \approx 0.0672 \).
05

Calculate Probability of Being Between 1.0 and 3.0

The probability that the path length is between \(1.0\) and \(3.0\) is \( P(1 < X \leq 3) = P(X \leq 3) - P(X \leq 1) \). Using the CDF formula, \( P(X \leq x) = 1 - e^{-\lambda x} \), we find \( P(1 < X \leq 3) = (1 - e^{-2.79}) - (1 - e^{-0.93}) \approx 0.8485 - 0.5935 = 0.2550 \).
06

Find 90th Percentile (Value Exceeded by 10%)

The value that is exceeded by only 10% of all path lengths is the 90th percentile. For the exponential distribution, the percentile is given by solving \( 1 - e^{-\lambda x} = 0.90 \) which gives \( e^{-\lambda x} = 0.10 \). Solving for \( x \), we find \( x = -\frac{\ln(0.10)}{0.93} \approx 2.481 \mu m \).

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.

Expected Value
In the context of the exponential distribution, the expected value represents the average path length a photon might have when its path length follows this distribution. The expected value allows us to understand what a typical path might look like under these circumstances. It is crucial for interpreting average outcomes and making predictions.In an exponential distribution with parameter \( \lambda \), the expected value \( E(X) \) is calculated as \( E(X) = \frac{1}{\lambda} \). Plugging in the given \( \lambda = 0.93 \), we find:- \( E(X) = \frac{1}{0.93} \approx 1.075 \mu m \)This means, on average, a photon will travel approximately 1.075 micrometers. This calculation is essential for scientists and engineers working on light behavior through mediums.
Standard Deviation
Standard deviation is a measure of the spread or variability around the expected value within a distribution, giving insight into the range of possible values the path length might take.For the exponential distribution, the standard deviation \( \sigma \) is the same as the expected value: \( \sigma = \frac{1}{\lambda} \). This is unique to the exponential distribution.- Thus, with \( \lambda = 0.93 \):- \( \sigma = \frac{1}{0.93} \approx 1.075 \mu m \)The identical value for both expected value and standard deviation indicates that the spread of the data around the mean is sizable, making it an crucial component in predicting occurrences and preparing for variability. Understanding this concept is vital to assessing risk and reliability within experiments.
Probability Calculations
Probability calculations allow us to determine how likely it is for a variable within an exponential distribution to occur in a specific range or exceed particular values.
  • **Probability of Exceeding 3.0**: Calculated using \( P(X > 3) = e^{-0.93 \times 3} \approx 0.0672 \). This tells us there's about a 6.72% chance a photon's path will exceed 3 micrometers.
  • **Probability Between 1.0 and 3.0**: Found by \( P(1 < X \leq 3) = (1 - e^{-2.79}) - (1 - e^{-0.93}) \approx 0.2550 \). So, there's a 25.50% probability the path length falls between 1 and 3 micrometers.
These probabilities provide a clearer understanding of path length distribution, crucial for making informed predictions and assessments in the field of photoresist research.
Percentiles
Percentiles are critical as they convey the value below which a specific percentage of observations fall within a distribution. In exponential distributions, percentiles are computed by evaluating the cumulative distribution function (CDF).For example, to find the 90th percentile, which is the value exceeded by only 10% of all the path lengths, we solve the equation:- \( 1 - e^{-0.93x} = 0.90 \)This simplifies to:- \( e^{-0.93x} = 0.10 \),- Solving for \( x \), we find \( x = -\frac{\ln(0.10)}{0.93} \approx 2.481 \mu m \).This means only 10% of photons travel more than approximately 2.481 micrometers. These percentile calculations are invaluable for researchers in determining threshold values and for setting limits and expectations in experimental designs.

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

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 ?

57\. a. Show that if \(X\) has a normal distribution with parameters \(\mu\) and \(\sigma\), then \(Y=a X+b\) (a linear function of \(X\) ) also has a normal distribution. What are the parameters of the distribution of \(Y\) [i.e., \(E(Y)\) and \(V(Y)]\) ? [Hint: Write the cdf of \(Y, P(Y \leq y)\), as an integral involving the pdf of \(X\), and then differentiate with respect to \(y\) to get the pdf of \(Y\).] b. If, when measured in \({ }^{\circ} \mathrm{C}\), temperature is normally distributed with mean 115 and standard deviation 2 , what can be said about the distribution of temperature measured in \({ }^{\circ} \mathrm{F}\) ?

The weight distribution of parcels sent in a certain manner is normal with mean value \(12 \mathrm{lb}\) and standard deviation \(3.5 \mathrm{lb}\). The parcel service wishes to establish a weight value \(c\) beyond which there will be a surcharge. What value of \(c\) is such that \(99 \%\) of all parcels are at least \(1 \mathrm{lb}\) under the surcharge weight?

When circuit boards used in the manufacture of compact dise players are tested, the long-run percentage of defectives is \(5 \%\). Suppose that a batch of 250 boards has been received and that the condition of any particular board is independent of that of any other board. a. What is the approximate probability that at least \(10 \%\) of the boards in the batch are defective? b. What is the approximate probability that there are exactly 10 defectives in the batch?

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.

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.