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Snowcat [4.5K]
3 years ago
9

Help with this question

Mathematics
1 answer:
jonny [76]3 years ago
3 0

Answer: 50%

Step-by-step explanation: 180/390 = 1/2 =0.5

0.5*100 = 50%

No questions asked

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Answer:

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

In which

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Over a long period of time, an average of 14 particles per minute occurs. Assume the arrival of particles at the counter follows a Poisson distribution. Find the probability that at least one particle arrives in a particular one second period.

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Either no particle arrives, or at least one does. The sum of the probabilities of these events is decimal 1. So

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We want P(X \geq 1). So

P(X \geq 1) = 1 - P(X = 0)

In which

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P(X = 0) = \frac{e^{-0.2333}*(0.2333)^{0}}{(0)!} = 0.7919

P(X \geq 1) = 1 - P(X = 0) = 1 - 0.7919 = 0.2081

0.2081 = 20.81% probability that at least one particle arrives in a particular one second period.

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