Sarah, a psychiatrist, receives a client according to a Poisson distribution on average once every $15$ minutes. Sarah is attempting to rest in between the client arrival. She is going to set an alarm so that the probability of a client arriving in that time is less than $40\%$. How long can Sarah set her alarm for?
My solution:
Let $N(t)$ be the number of patients arriving in t minutes.
Therefore:
$$N(t) \sim Poisson\left(λ = \frac t{15}\right)$$
The probability of at least one patient arriving in $t$ minutes is:
$$Pr(N(t \ge 1)) = 1 - Pr(N(t = 0)) = 1-e^{\frac{-t/15}{15}}$$
Therefore,
$$Pr(N(t \ge 1) \le 0.40\\1-e^{\frac{-t/15}{15}}\le 0.40\\t \le -30 \log(3) + 30i\pi n\text{ and }n\in\mathbb Z$$
Therefore, Sarah can set her alarm for or less than $-30 \log(3) + 30i\pi n$ minutes for $n\in\mathbb Z$.
Is this correct?
I've found the answer! Thanks for the hints!
My solution:
Let $N(t)$ be the number of patients arriving in t minutes.
Therefore:
$$N(t) \sim Poisson\left(λ = \frac t{15}\right)$$
The probability of at least one patient arriving in $t$ minutes is: $$Pr(N(t)\le 1)=1-Pr(N(t = 0))=1-e^{\frac{-t/15}{15}},$$
Therefore,
$$Pr(N(t \ge 1) \lt 0.40\\1-e^{\frac{-t/15}{15}}\lt 0.40\\e^{\frac{-t/15}{15}}\lt 0.60\\\frac{-t}{15}\left(\frac{ln(e)}{ln(e)}\right)\lt\left(\frac{ln(0.60)}{ln(e)}\right)\\t\lt -15(-0.5108)\\t\lt7.6624 $$
Therefore, Sarah can set her alarm for less than 7.6624 minutes such that the probability of a client arriving in that time is less than 40%.