We are given that an event is expected to occur 20 times a day

Question

We are given that an event is expected to occur 20 times a day. I think that this event is a continuous random variable, and there should be really no effect of these events on one another. How would I start to find the probability that this event occurs only 15 or fewer times on a given day and how would I find the probability between different times of these events occuring? From what I've gathered so far there will be roughly 1.2 hours between events by using E(T) = 1/lambda. I'd prefer not to be given answers and just the best way to start this problem out, I've been trying to find the formulas I would need through google but have only been able to find homework assignments from certain colleges.

Thank you!

Answer 1

Here are indications of the arguments that lead to answers for both of you questions.

Probability of 15 or fewer in a day. If $X \sim \mathsf{Pois}(\lambda = 20)$ is the number of events per day, then $$P(X \le 15) = \sum_{x=0}^{15} \frac{e^{-20}20^x}{x!} = 0.1565,$$ evaluated using R statistical software as:

 sum(dpois(0:15, 20))  # 'dpois' is Poisson PDF
 ## 0.1565131

 ppois(15, 20)         # 'ppois' is Poisson CDF
 ## 0.1565131

This may be roughly approximated by a normal distribution: $$P(X \le 15) = P(X < 15.5) = P\left(\frac{X-\lambda}{\sqrt{\lambda}} < \frac{15.5-20}{\sqrt{20}}\right) \approx P(Z < -1.006) = 0.1572,$$ where $Z$ is standard normal, and the probability can be found from printed tables.

Expected interarrival time. Starting at time $t = 0,$ the probability that the waiting time $W$ for the first event exceeds $t$ is $P(W > t)$ is the same as the probability $P(X_t = 0)$ of no event in $(0, t),$ where $X_t \sim \mathsf{Pois}(\lambda_t = t\lambda).$

Then $P(W > t) = P(X_t = 0) = e^{-\lambda t},$ so that the CDF of $W$ is $F_W(t) = P(W \le t) = 1 - e^{-\lambda t},$ for $t > 0.$ Then by differentiation, the density function of $W$ is $f_W(t) = \lambda e^{-\lambda t},$ for $t > 0.$ This is the density function of $W \sim \mathsf{Exp}(rate = \lambda).$ And $E(W) = 1/\lambda$ can be found using integration by parts in $E(W) = \int_0^\infty tf_W(t)\,dt.$

By the no-memory property of the exponential distribution the waiting time for the first event starting at $t=0$ is the same as the waiting time for the $(i+1)$st event starting at the time of the $i$th event.