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11 November, 04:20

The probability density function of the time to failure of an electronic component in a copier (in hours) is f (x) = (e^-x/1076) / (1076) for x>0. Determine the probability that

A) A component lasts more than 3000 hours before failure. (Round the answer to 3 decimal places.)

B) A component fails in the interval from 1000 to 2000 hours. (Round the answer to 3 decimal places.)

C) A component fails before 1000 hours. (Round the answer to 3 decimal places.)

D) Determine the number of hours at which 10% of all components have failed. (Round the answer to the nearest integer.)

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Answers (1)
  1. 11 November, 04:44
    0
    (a) 0.06154

    (b) 0.2389

    (c) 0.6052

    (d) 2478

    Step-by-step explanation:

    probability density function of the time to failure of an electronic component in a copier (in hours) is

    P (x) = 1/1076e^-x/1076

    λ = 1/1076

    A) A component lasts more than 3000 hours before failure:

    P (x>3000) = 1 - e^-3000/1076

    = 0.06154

    B) A component fails in the interval from 1000 to 2000 hours:

    P (1000>x>2000) = 1 - e^-2000/1076 - 1 + e^-1000/1076 = e^-1000/1076 - e^-2000/1076 = 0.3948 - 0.1559

    = 0.2389

    C) A component fails before 1000 hours:

    P (x<1000) = 0.6052

    D) The number of hours at which 10% of all components have failed:

    e^-x/1076 = 0.1

    = - x/1076

    = ln (0.1)

    x = (2.3026) * (1076)

    x = 2478
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