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Chapter 5: Problem 87

In a food processing and packaging plant, there are, on the average, two packaging machine breakdowns per week. Assume the weekly machine breakdowns follow a Poisson distribution. a. What is the probability that there are no machine breakdowns in a given week? b. Calculate the probability that there are no more than two machine breakdowns in a given week.

Short Answer

Expert verified
Answer: a) The probability of no machine breakdowns in a given week is approximately 13.53%. b) The probability of no more than two machine breakdowns in a given week is approximately 67.67%.

Step by step solution

01

Calculate the probability of no machine breakdowns in a given week (k=0)

Since 位=2 and k=0, we can plug these values into the Poisson Probability Formula: P(X=0) = (2^0 * e^(-2)) / 0! = (1 * e^(-2)) / 1 = e^(-2) Now, let's calculate e^(-2) using a calculator: P(X=0) 鈮� 0.1353 So, the probability of no machine breakdowns in a given week is approximately 13.53%.
02

Calculate the probability of no more than two machine breakdowns in a given week (k=0, 1, or 2)

We need to find the probabilities for k=0, k=1, and k=2, and then sum them up: P(X=0, 1, or 2) = P(X=0) + P(X=1) + P(X=2) We have already calculated P(X=0) in Step 1. Now let's calculate P(X=1) and P(X=2): P(X=1) = (2^1 * e^(-2)) / 1! = (2 * e^(-2)) / 1 鈮� 0.2707 P(X=2) = (2^2 * e^(-2)) / 2! = (4 * e^(-2)) / 2 鈮� 0.2707 Now, summing them up: P(X=0, 1, or 2) 鈮� 0.1353 + 0.2707 + 0.2707 鈮� 0.6767 So, the probability of no more than two machine breakdowns in a given week is approximately 67.67%.

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