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The average number of surface defects in 20meter square of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40meter square of paper through random selection? (Use Poisson Distribution) Ops: A. 0.062 B. 0.041 C. 0.059 D. 0.031?
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The average number of surface defects in 20meter square of paper produ...
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Poisson Distribution is used to find the probability of a given number of events occurring in a fixed interval of time or space, assuming that these events occur independently and at a constant rate.
Given that the average number of defects in 20m^2 of paper is 3, we can assume that the defects occur at a rate of λ = 3/20 defects per square meter.
We need to find the probability of finding no more than 2 defects in 40m^2 of paper through random selection.
Let X be the number of defects in 40m^2 of paper. Then X follows a Poisson distribution with parameter λ' = 3/20 * 40 = 6.
To find the required probability, we can use the Poisson probability mass function:
P(X ≤ 2) = Σ (e^-λ' * λ'^k / k!) for k = 0 to 2
P(X ≤ 2) = e^-6 * (6^0 / 0!) + e^-6 * (6^1 / 1!) + e^-6 * (6^2 / 2!)
P(X ≤ 2) = 0.0025 + 0.0149 + 0.0447
P(X ≤ 2) = 0.0621
Therefore, the probability of finding no more than 2 defects in 40m^2 of paper through random selection is 0.0621.
Answer: A. 0.062
Poisson Distribution is used to find the probability of a given number of events occurring in a fixed interval of time or space, assuming that these events occur independently and at a constant rate.
Given that the average number of defects in 20m^2 of paper is 3, we can assume that the defects occur at a rate of λ = 3/20 defects per square meter.
We need to find the probability of finding no more than 2 defects in 40m^2 of paper through random selection.
Let X be the number of defects in 40m^2 of paper. Then X follows a Poisson distribution with parameter λ' = 3/20 * 40 = 6.
To find the required probability, we can use the Poisson probability mass function:
P(X ≤ 2) = Σ (e^-λ' * λ'^k / k!) for k = 0 to 2
P(X ≤ 2) = e^-6 * (6^0 / 0!) + e^-6 * (6^1 / 1!) + e^-6 * (6^2 / 2!)
P(X ≤ 2) = 0.0025 + 0.0149 + 0.0447
P(X ≤ 2) = 0.0621
Therefore, the probability of finding no more than 2 defects in 40m^2 of paper through random selection is 0.0621.
Answer: A. 0.062
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The average number of surface defects in 20meter square of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40meter square of paper through random selection? (Use Poisson Distribution) Ops: A. 0.062 B. 0.041 C. 0.059 D. 0.031?
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The average number of surface defects in 20meter square of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40meter square of paper through random selection? (Use Poisson Distribution) Ops: A. 0.062 B. 0.041 C. 0.059 D. 0.031? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about The average number of surface defects in 20meter square of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40meter square of paper through random selection? (Use Poisson Distribution) Ops: A. 0.062 B. 0.041 C. 0.059 D. 0.031? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average number of surface defects in 20meter square of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40meter square of paper through random selection? (Use Poisson Distribution) Ops: A. 0.062 B. 0.041 C. 0.059 D. 0.031?.
The average number of surface defects in 20meter square of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40meter square of paper through random selection? (Use Poisson Distribution) Ops: A. 0.062 B. 0.041 C. 0.059 D. 0.031? for Computer Science Engineering (CSE) 2024 is part of Computer Science Engineering (CSE) preparation. The Question and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus. Information about The average number of surface defects in 20meter square of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40meter square of paper through random selection? (Use Poisson Distribution) Ops: A. 0.062 B. 0.041 C. 0.059 D. 0.031? covers all topics & solutions for Computer Science Engineering (CSE) 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The average number of surface defects in 20meter square of paper produced by a process is 3. What is the probability of finding no more than 2 defects in 40meter square of paper through random selection? (Use Poisson Distribution) Ops: A. 0.062 B. 0.041 C. 0.059 D. 0.031?.
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