0
$\begingroup$

So I have a question that basically says: Customers arrive at a facility at a mean rate of $12$ per minute. Find the Probability that a gap of greater than $7$ seconds occurs between two consecutive arrivals.

I've started the question and found the mean time between arrivals ($\frac{1}{12}$ minutes $= 5$ seconds). I also know I need to use the negative exponential distribution, I should use $e^{-λt}$, however, I'm not sure what my $\lambda$ and t value should be. I think it might be quite simple and I'm overthinking it, any help on how I would obtain $\lambda$ and t would be great.

Thanks in advance

$\endgroup$
1

1 Answer 1

2
$\begingroup$

If your time was in unit of minutes, your $\lambda$ would be $1/12$, but it is not, right, it is more convenient for us to use seconds and use $\lambda = 0.2$ per second. Then we use the cumulative distribudion function to find the probability of being outside the interval $[0, 7]$ seconds. So we may take $1 - $P(being in that interval).

The CDF of $\lambda e^{- \lambda t}$ is $1 - e^{-\lambda t}$.

This equals $1 - CDF[7]$ with the right $\lambda$ (rate, intensity).

$$ 1 - CDF[7] = 1 - (1 - e^{-\lambda 7} )= e^{-0.2*7} = 0.247$$

$\endgroup$
4
  • $\begingroup$ Thanks, this was really useful. Just wondering, because I got the same answer just now, would it be easier to do your way or to do: P(t>7)=e^-(12 x (7/60)) $\endgroup$
    – Maximus
    Commented Mar 3, 2021 at 14:37
  • $\begingroup$ Ok, not is it easier but does this way work as well as yours? (Sorry not sure how to word my question properly) $\endgroup$
    – Maximus
    Commented Mar 3, 2021 at 14:37
  • $\begingroup$ Yes, that is equivalent. $\endgroup$ Commented Mar 3, 2021 at 14:38
  • $\begingroup$ oh ok, great, thanks a lot $\endgroup$
    – Maximus
    Commented Mar 3, 2021 at 14:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .