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8_murik_8 [283]
2 years ago
11

On average, 2.8 babies are born each day at a local hospital. Assuming the Poisson distribution, what is the probability that no

babies are born today?
Mathematics
1 answer:
Anon25 [30]2 years ago
4 0

Answer:

The probability is P(X = 0) =  0.6997

Step-by-step explanation:

From the question we are told that

   The  mean is  \mu =  2.8

Generally the Poisson distribution constant  \lambda is mathematically represented as

        \lambda = \frac{1}{\mu }

=>     \lambda = \frac{1}{2.8 }

=>     \lambda = 0.3571

Generally the probability distribution for Poisson distribution is mathematically represented as  

     P(X = x) =  \frac{(\lambda t)^x e^{-t \lambda}}{x!}

Here  t =  1 day

Generally the probability  that no babies are born today is mathematically represented as

       P(X = 0) =  \frac{(0.3571 * 1 )^0 e^{-1 * 0.3571}}{0!}

=>    P(X = 0) =  0.6997

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