# Normal interarrival times

I have the interarrival time $$t$$ follows a normal distribution of $$N(8,4)$$. I'm trying to find

1. $$P(t < 0)$$
2. The probability that the 16th and the 9th customer arrive within 55 minutes of each other

For 1, it would be this right?

$$P(t < 0) = \int_{-\infty}^{0} \frac{1}{2\sqrt{2\pi}} e^{-(t-8)^2/(2 * 2^2)}$$

For 2, I'm a bit confused. Is this a Poisson distribution then? Because t already follows a Normal distribution.

Can I have some helps with 1 and 2 please?

• In general, if $t\sim\mathcal{N}\left(\mu, \sigma^2\right)$ where $\sigma > 0$, then by standardising, we have $P(t < a) = P\left(Z < \frac{a-\mu}{\sigma}\right) =\Phi\left(\frac{a-\mu}{\sigma}\right)$, where $Z$ is a standard normal random variable and $\Phi$ is the standard normal CDF. If you have a table of standard normal CDF values, you can use this to calculate this expression. Your integral expression is also correct, but how are you expected to calculate it? Mar 7 '19 at 22:22
• Isn't an interarrival time a non -negative random variable? How can it be normally distributed?. Mar 7 '19 at 23:30
• @MinusOne-Twelfth Oh I was going to use the CDF for that but I thought I can only use it if I were to calculate $P(t \le 0)$. This is me not really knowing CDF and PDF as well.
– PTN
Mar 8 '19 at 0:17
• Possibly "for simplicity", we are modelling the interarrival time as normally distributed (maybe we want to model it as peaked around $8$ minutes for example), and maybe the question asking for $P(t < 0)$ at the start is asking us how likely it is that under this model, we get an interarrival time being "absurd" (negative); if this probability turns out to be very small, then it is perhaps not too much of a concern that we can get negative times from this model. That's my best guess, anyway. Mar 8 '19 at 0:51
• @PTN We can use the CDF for $\le$ and also for $<$ (since we are dealing with a continuous random variable here, the answer is the same either way). Mar 8 '19 at 0:52

Part 1 looks right to me. I would guess that Part 2 is asking for the probability that the wait times $$t_{10:16} = \sum_{i=10}^{16} t_i$$ are less than 55 minutes, that is $$P(t_{10:16} < 55)$$. Assuming that the arrival times are independent, then $$t_{10:16}$$ is also a Gaussian $$N(7 \cdot 8, 7 \cdot 4)$$ and hence we can find $$P(t_{10:16} < 55)$$ just as in part 1.
• That makes sense. I just have questions at how did you get the $7$, and why is it $t_{10:16}$ and not $t_{9:16}$
• The 7 is because there are 7 terms in the sum (we count starting at 10). I used $t_{10:16}$ because $t_9$ is irrelevant -- it is the waiting time for the 9th person to arrive, but the second problems asks us to start counting when the 9th person arrives. Mar 8 '19 at 23:02