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Jobisdone [24]
3 years ago
14

A local animal rescue organization receives an average of 0.55 rescue calls per hour. Use the Poisson distribution to find the p

robability that during a randomly selected hour, the organization will receive fewer than two calls.A) 0.087
B) 0.894
C) 0.317
D) 0.106
Mathematics
1 answer:
andreyandreev [35.5K]3 years ago
5 0

Answer:

B) 0.894

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.

A local animal rescue organization receives an average of 0.55 rescue calls per hour.

This means that \mu = 0.55

Probability that during a randomly selected hour, the organization will receive fewer than two calls.

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

In which

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

P(X = 0) = \frac{e^{-0.55}*(0.55)^{0}}{(0)!} = 0.577

P(X = 1) = \frac{e^{-0.55}*(0.55)^{1}}{(1)!} = 0.317

P(X < 2) = P(X = 0) + P(X = 1) = 0.577 + 0.317 = 0.894

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