Consider a Poisson distribution with μ = 6. a. Write the appropriate Poisson probability function. b. Compute f (2).

Question

Consider a Poisson distribution with μ = 6.

a. Write the appropriate Poisson probability function.
b. Compute f (2).
c. Compute f (1).
d. Compute P(x≥2)

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King 5 years 2021-08-29T23:51:31+00:00 1 Answers 59 views 0

Answers ( )

  1. Sapo1
    0
    2021-08-29T23:52:35+00:00
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    Answer:

    a) P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

    b) f(2) = 0.04462

    c) f(1) = 0.01487

    d) P(X \geq 2) = 0.93803

    Step-by-step explanation:

    In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

    P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}

    In which

    x is the number of successes

    e = 2.71828 is the Euler number

    \mu is the mean in the given interval.

    In this question:

    \mu = 6

    a. Write the appropriate Poisson probability function.

    Considering \mu = 6

    P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

    b. Compute f (2).

    This is P(X = 2). So

    P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

    P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

    So f(2) = 0.04462

    c. Compute f (1).

    This is P(X = 1). So

    P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

    P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

    So f(1) = 0.01487.

    d. Compute P(x≥2)

    This is:

    P(X \geq 2) = 1 - P(X < 2)

    In which:

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

    P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}

    P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248

    P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487

    P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462

    Then

    P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197

    P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803

    So

    P(X \geq 2) = 0.93803

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