Question

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?

Answer 1

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

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

=>    $P(X = 0) = 0.6997$