A random variable X has a Poisson distribution with a mean of 3. What is the probability that X P(1≤X≤3) ?.

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A random variable X has a Poisson distribution with a mean of 3. What is the probability that X P(1≤X≤3) ?.

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Thiên Hương 4 years 2021-07-19T03:50:23+00:00 1 Answers 15 views 0

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  1. Edana
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    2021-07-19T03:51:36+00:00
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    Answer:

    P(1≤X≤3) = 0.5974

    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 sucesses

    e = 2.71828 is the Euler number

    \mu is the mean in the given interval.

    Mean of 3

    This means that \mu = 3

    P(1≤X≤3) ?

    P(1 \leq X \leq 3) = P(X = 1) + P(X = 2) + P(X = 3)

    So

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

    P(X = 1) = \frac{e^{-3}*3^{1}}{(1)!} = 0.1494

    P(X = 2) = \frac{e^{-3}*3^{2}}{(2)!} = 0.2240

    P(X = 3) = \frac{e^{-3}*3^{3}}{(3)!} = 0.2240

    So

    P(1 \leq X \leq 3) = P(X = 1) + P(X = 2) + P(X = 3) = 0.1494 + 0.2240 + 0.2240 = 0.5974

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