Question

The number of machine breakdowns in a month follows a Poisson distribution with a mean of 3. An insurance contract covers repairs to the machine if more than N breakdowns occur in a month. N is selected to be the highest number such that the contract will provide repairs in at least 25% of the months. Determine N.

Answer 1

Answer:

N= 3

Step-by-step explanation:

given data

mean $\lambda$ =  3

provide repairs at least =  25%

solution

as here for poisson distribution

P(X=k) = $\frac{(\lambda )^ke^{-\lambda }}{k!}$   ......................1

so here

P(X>n) = 0.25

and

P(X>n) = $e^{-3 } \sum_{X=n}^{\infty } \frac{(3 )^k}{k!}\geq 0.25$      ..............2

so

P(X>3) = 0.3528

and

P(X>4) = 0.1847

so here  

N= 3