cal inspection system is used to distinguish among different part types. The probability of a correct classification of any part is ppose that three parts are inspected and that the classifications are independent. Let the random variable X denote the of parts that are correctly classified. Determine the probability mass function of X. our answers to four decimal places (e.g. 98.7654). f(x)
cal inspection system is used to distinguish among different part types. The probability of a correct classification of any part is ppose that three parts are inspected and that the classifications are independent. Let the random variable X denote the of parts that are correctly classified. Determine the probability mass function of X. our answers to four decimal places (e.g. 98.7654). f(x)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
2) Please provide a correct answer with a full method explanation and do not copy and paste from other similar questions.
![**Understanding Probability Mass Function (PMF) in Optical Inspection Systems**
An optical inspection system is utilized to differentiate between various part types. The probability that a part is correctly classified by this system is 0.91. Considering that three parts are inspected and their classifications are independent, we define the random variable \( X \) as the number of parts that are correctly classified.
To determine the probability mass function (PMF) of \( X \), use the following steps:
**Given:**
- Probability of a correct classification (success) for any part, \( p = 0.91 \)
- Number of parts inspected, \( n = 3 \)
The PMF \( f(x) \) for a binomial random variable, denoted as \( X \), is given by:
\[
f(x) = \binom{n}{x} p^x (1 - p)^{n - x}
\]
where \(\binom{n}{x}\) is the binomial coefficient calculated by \(\frac{n!}{x!(n - x)!}\).
**Calculate the PMF values and round your answers to four decimal places:**
\( x \): Number of parts correctly classified
\( f(x) \): Probability mass function of \( X \)
| \( x \) | \( f(x) \) |
|------|-------|
| 0 | |
| 1 | |
| 2 | |
| 3 | |
Fill in the probabilities \( f(0) \), \( f(1) \), \( f(2) \), and \( f(3) \) the binomial formula:
1. For \( x = 0 \):
\[ f(0) = \binom{3}{0} 0.91^0 (1 - 0.91)^3 \]
2. For \( x = 1 \):
\[ f(1) = \binom{3}{1} 0.91^1 (1 - 0.91)^2 \]
3. For \( x = 2 \):
\[ f(2) = \binom{3}{2} 0.91^2 (1 - 0.91)^1 \]
4. For \( x = 3 \):
\[ f(3) = \binom{3}{3}](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F7824f8d0-65fb-442c-84f7-7ea5547717e1%2Facaf83a7-8db0-4429-a53f-828eb29d645d%2F24zx8gu_processed.png&w=3840&q=75)
Transcribed Image Text:**Understanding Probability Mass Function (PMF) in Optical Inspection Systems**
An optical inspection system is utilized to differentiate between various part types. The probability that a part is correctly classified by this system is 0.91. Considering that three parts are inspected and their classifications are independent, we define the random variable \( X \) as the number of parts that are correctly classified.
To determine the probability mass function (PMF) of \( X \), use the following steps:
**Given:**
- Probability of a correct classification (success) for any part, \( p = 0.91 \)
- Number of parts inspected, \( n = 3 \)
The PMF \( f(x) \) for a binomial random variable, denoted as \( X \), is given by:
\[
f(x) = \binom{n}{x} p^x (1 - p)^{n - x}
\]
where \(\binom{n}{x}\) is the binomial coefficient calculated by \(\frac{n!}{x!(n - x)!}\).
**Calculate the PMF values and round your answers to four decimal places:**
\( x \): Number of parts correctly classified
\( f(x) \): Probability mass function of \( X \)
| \( x \) | \( f(x) \) |
|------|-------|
| 0 | |
| 1 | |
| 2 | |
| 3 | |
Fill in the probabilities \( f(0) \), \( f(1) \), \( f(2) \), and \( f(3) \) the binomial formula:
1. For \( x = 0 \):
\[ f(0) = \binom{3}{0} 0.91^0 (1 - 0.91)^3 \]
2. For \( x = 1 \):
\[ f(1) = \binom{3}{1} 0.91^1 (1 - 0.91)^2 \]
3. For \( x = 2 \):
\[ f(2) = \binom{3}{2} 0.91^2 (1 - 0.91)^1 \]
4. For \( x = 3 \):
\[ f(3) = \binom{3}{3}
Expert Solution
This question has been solved!
Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step by step
Solved in 2 steps with 2 images
Recommended textbooks for you
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
MATLAB: An Introduction with Applications
Statistics
ISBN:
9781119256830
Author:
Amos Gilat
Publisher:
John Wiley & Sons Inc
Probability and Statistics for Engineering and th…
Statistics
ISBN:
9781305251809
Author:
Jay L. Devore
Publisher:
Cengage Learning
Statistics for The Behavioral Sciences (MindTap C…
Statistics
ISBN:
9781305504912
Author:
Frederick J Gravetter, Larry B. Wallnau
Publisher:
Cengage Learning
Elementary Statistics: Picturing the World (7th E…
Statistics
ISBN:
9780134683416
Author:
Ron Larson, Betsy Farber
Publisher:
PEARSON
The Basic Practice of Statistics
Statistics
ISBN:
9781319042578
Author:
David S. Moore, William I. Notz, Michael A. Fligner
Publisher:
W. H. Freeman
Introduction to the Practice of Statistics
Statistics
ISBN:
9781319013387
Author:
David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:
W. H. Freeman