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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 0.002. 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 p = 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 (in the batch of 1000) needing repair within their warranty periods equals . (Express this value precise to three decimal places, e.g., 0.281, 0.335, 0.721.)

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 0.002. 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 p = 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 (in the batch of 1000) needing repair within their warranty periods equals . (Express this value precise to three decimal places, e.g., 0.281, 0.335, 0.721.)

Calculus For The Life Sciences
Calculus For The Life Sciences
2nd Edition
ISBN:9780321964038
Author:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Publisher:GREENWELL, Raymond N., RITCHEY, Nathan P., Lial, Margaret L.
Chapter13: Probability And Calculus
Section13.2: Expected Value And Variance Of Continuous Random Variables
Problem 10E
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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 0.002. 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 p = 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 (in the batch of 1000) needing repair within their warranty periods equals . (Express this value precise to three decimal places, e.g., 0.281, 0.335, 0.721.)
Transcribed Image Text: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 0.002. 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 p = 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 (in the batch of 1000) needing repair within their warranty periods equals . (Express this value precise to three decimal places, e.g., 0.281, 0.335, 0.721.)
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