LCD display screens are fabricated using a process which averages 6.5 faulty pixels per square centimeter. Suppose a small screen used for a clock has an area of 1 square centimeter.Use the Poisson distribution to find the probability that the screen has 2 or fewer faulty pixels.Do not round intermediate computations, and round your answer to three decimal places.(If necessary, consult a list of formulas.)
Question
LCD display screens are fabricated using a process which averages 6.5 faulty pixels per square centimeter. Suppose a small screen used for a clock has an area of 1 square centimeter.Use the Poisson distribution to find the probability that the screen has 2 or fewer faulty pixels.Do not round intermediate computations, and round your answer to three decimal places.(If necessary, consult a list of formulas.)
Solution
The Poisson distribution is a probability distribution that can be used to model the number of events (in this case, the number of faulty pixels) in a fixed area or volume. The formula for the Poisson distribution is:
P(X=k) = (λ^k * e^-λ) / k!
where:
- P(X=k) is the probability of k events occurring in an interval,
- λ is the average rate of value (in this case, 6.5 faulty pixels per square centimeter),
- e is the base of the natural logarithm (approximately equal to 2.71828),
- k! is the factorial of k.
We want to find the probability that the screen has 2 or fewer faulty pixels, so we need to calculate P(X=0), P(X=1), and P(X=2), and then add these probabilities together.
-
Calculate P(X=0): P(X=0) = (6.5^0 * e^-6.5) / 0! = e^-6.5 = 0.0015
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Calculate P(X=1): P(X=1) = (6.5^1 * e^-6.5) / 1! = 6.5 * e^-6.5 = 0.0097
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Calculate P(X=2): P(X=2) = (6.5^2 * e^-6.5) / 2! = 21.125 * e^-6.5 = 0.0316
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Add the probabilities together: P(X<=2) = P(X=0) + P(X=1) + P(X=2) = 0.0015 + 0.0097 + 0.0316 = 0.0428
So, the probability that the screen has 2 or fewer faulty pixels is 0.0428, or 4.28% when rounded to three decimal places.
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