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Anastaziya [24]
3 years ago
13

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

Mathematics
1 answer:
frosja888 [35]3 years ago
4 0

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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