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

The probability is
`P(X = 0) =  0.6997`

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 = (1)/(\mu )`

=>
`\lambda = (1)/(2.8 )`

=>
`\lambda = 0.3571`

Generally the probability distribution for Poisson distribution is mathematically represented as

`P(X = x) =  ((\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) =  ((0.3571 * 1 )^0 e^(-1 * 0.3571))/(0!)`

=>
`P(X = 0) =  0.6997`