Question

Last month, your company sold 1,000 new watches Past experience indicates that the probability that a new watch will need repair during its warranty period is 0002 Let X be the number of watches among the 1,000 that will need repair during its warranty period. X is binomially distributed with n = 1.000 and o-0.002 The distribution of X, however, can be approximated using a distribution with mean equal to Using this approximating distribution, we find that the probability of having at most two watches on the batch of 1000) needing repair within their warranty periods quals (Express this value precise to three decimal places, es 0.2810.335.0.721)

Answer 1

The probability of having at most two watches needing repair within their warranty periods can be approximated using the Poisson distribution. The mean of the approximating Poisson distribution is 2. The probability of having at most two watches needing repair is approximately 0.676.

Step-by-step explanation:

The probability of having at most two watches needing repair within their warranty periods can be calculated using the binomial distribution. Let's define X as the number of watches that need repair. X follows a binomial distribution with n = 1,000 and p = 0.002 (probability of a watch needing repair). To approximate this distribution, we can use the Poisson distribution.

The mean of the approximating Poisson distribution is given by µ = n * p = 1,000 * 0.002 = 2. Therefore, we can approximate X using a Poisson distribution with a mean of 2.

To find the probability of having at most two watches needing repair, we can sum up the probabilities of X = 0, 1, and 2 using the Poisson distribution formula:

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

Calculating this probability using the Poisson distribution, we find that P(X ≤ 2) ≈ 0.676