Suppose X has an exponential distribution with mean equal to 23. Determine the following: (a) P(X >10) (b) P(X >20)

Question

Suppose X has an exponential distribution with mean equal to 23. Determine the following:
(a) P(X >10)
(b) P(X >20)
(c) P(X <30)
(d) Find the value of x such that P(X

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Xavia 5 years 2021-08-19T12:41:20+00:00 1 Answers 328 views 1

Answers ( )

  1. thongdat2
    0
    2021-08-19T12:42:24+00:00
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    Answer:

    a) P(X > 10) = 0.6473

    b) P(X > 20) = 0.4190

    c) P(X < 30) = 0.7288

    d) x = 68.87

    Step-by-step explanation:

    Exponential distribution:

    The exponential probability distribution, with mean m, is described by the following equation:

    f(x) = \mu e^{-\mu x}

    In which \mu = \frac{1}{m} is the decay parameter.

    The probability that x is lower or equal to a is given by:

    P(X \leq x) = \int\limits^a_0 {f(x)} \, dx

    Which has the following solution:

    P(X \leq x) = 1 - e^{-\mu x}

    The probability of finding a value higher than x is:

    P(X > x) = 1 - P(X \leq x) = 1 - (1 - e^{-\mu x}) = e^{-\mu x}

    Mean equal to 23.

    This means that m = 23, \mu = \frac{1}{23} = 0.0435

    (a) P(X >10)

    P(X > 10) = e^{-0.0435*10} = 0.6473

    So

    P(X > 10) = 0.6473

    (b) P(X >20)

    P(X > 20) = e^{-0.0435*20} = 0.4190

    So

    P(X > 20) = 0.4190

    (c) P(X <30)

    P(X \leq 30) = 1 - e^{-0.0435*30} = 0.7288

    So

    P(X < 30) = 0.7288

    (d) Find the value of x such that P(X > x) = 0.05

    So

    P(X > x) = e^{-\mu x}

    0.05 = e^{-0.0435x}

    \ln{e^{-0.0435x}} = \ln{0.05}

    -0.0435x = \ln{0.05}

    x = -\frac{\ln{0.05}}{0.0435}

    x = 68.87

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