The UCD Health Centre believe that the number of students who contract conjunctivitis follows a Poisson Process with parameter λ = 1/3 per month
The UCD Health Centre hopes to reduce services during the summer months when there are no undergraduate students on campus. The director of the UCD Health Centre says the services can be reduced if the probability of no students contracting conjunctivitis is at most 0.75. Assuming that no student has contracted conjunctivitis at the start of the summer, for how long can services be reduced?
So I had a method in my mind.
Integrate I = ∫[(lambda)^0*exp(-lambda)]/0! from 1 to n and this I≤ 0.75 and solve for n. Is that correct or should I use exponential distribution ?
The number in $t$ months has Poisson distribution paramter $(1/3)t$. So the probability the number of students is $0$ is $e^{-t/3}$. Set this equal to $0.75$ and solve for $t$.
Alternately, you could use the relationship between the Poisson and the exponential. The probability that the "waiting time" until the first case is $\gt t$ is $\int_{t}^\infty (1/3)e^{-x/3}\,dx$.