1

I need to find the solution of $\dfrac{4^n}{n!}=51.8682$. Though I need $n$ to be a natural number, if $n\in\mathbb{R}$, I can take the celling of $n$ for problem purpose.
It can be done by trial and error method. But one thing is noticeable that the denominator is increasing rapidly than the numerator. For $n=1,2,3$ the R.H.S is very far from the result. As $n$ increases, the fraction decreases and hence the fraction does not reach even at $50$.
Is there any way to solve this equation?

Question added A cafeteria with self-service has an arrival rate of $12$/hour. The average time taken by a person to collect and eat his meal is $20$ minutes. Assuming that the inter-arrival times are exponentially distributed, how many seats must the cafeteria have to accommodate each customer with probability $0.95$?

Solution: Here $\lambda=12$ persons/hour and $\mu=\dfrac{60}{20}=3$ customers/hour. Therefore $\rho=\dfrac {12}{3}=4$. We need to find $n$ so that $P_n=0.95$. Therefore we have $ \dfrac{\rho^ne^{-\rho}}{n!}=0.95\implies \dfrac{4^ne^{-4}}{n!}=0.95\implies \dfrac{4^n}{n!}=51.8682$

Manjoy Das
  • 1,006
  • 7
  • 21
  • 1
    As you noticed, starting with $n=4$ it becomes a decreasing sequence. Thus, there is no solution. – Vasili Oct 24 '22 at 19:37
  • @Vasili But there has to be a solution as the problem indicates. – Manjoy Das Oct 24 '22 at 19:38
  • 1
    The question concerns a cumulative probability. – lulu Oct 24 '22 at 19:46
  • @lulu would you like to explain it a bit more? – Manjoy Das Oct 24 '22 at 19:54
  • You are looking at the probability that exactly $n$ show up in some interval. You should be looking at the probability that at most $n$ people show up. – lulu Oct 24 '22 at 20:03
  • Note: the problem is a bit strange in that so few people come to the cafeteria. I wouldn't think you'd need very many seats to accommodate them. – lulu Oct 24 '22 at 20:05
  • @lulu Thanks for your suggestion. But I think in that case also, this problem would arise to find the value of $n$. – Manjoy Das Oct 24 '22 at 20:18
  • 1
    That is a trivial numerical problem. Approximations are possible, but why bother? The numbers are very small indeed so just do it exactly. – lulu Oct 24 '22 at 20:19
  • @lulu It is better not to use adjective "trivial". People who aren't at ease with maths receive the use of it as "you aren't intelligent enough to understand" – Jean Marie Oct 24 '22 at 21:21
  • 1
    @JeanMarie Thanks, fair point. To be clear, I intended it to read as "this is a standard numerical problem, and it is probably optimal to solve it that way" but I appreciate that it could have been interpreted more harshly. – lulu Oct 24 '22 at 21:25
  • @lulu I'm not finding how cumulative density would resolve this issue. If it's possible for you to answer this question rather than commenting, it would be very nice! – Manjoy Das Oct 24 '22 at 21:31
  • I don't understand. $X$, the number of people who arrive in a $20$ minute window, is a Poisson variable with mean $4$. We want the least $n$ such that $P(X≤n)≥.95$ so just compute the cumulative Poisson distribution for $n=0, 1, 2, \cdots$ until you find one that works. For large $n$, of course, $P(X≤n)$ approaches $1$ so we know there is a value that works and, indeed, it's quite small. – lulu Oct 24 '22 at 22:51

1 Answers1

1

Working in the real domain, you can have an almost exact explicit solution.

Making the problem more general, you want to solve for $n$ $$\frac {a^n}{n!}=b \qquad \implies \qquad n!=\frac {a^n}{b}$$

If you look at this old question of mine, you will see a superb approximation by @robjohn (one of our moderators).

Just make $10^k=\frac 1b$ that is to say $k=-\frac {\log(b)}{\log(10)}$ to get $$n \sim e a \exp\left(W\left( -\frac{\log \left(2 \pi a b^2\right)}{2 e a} \right)\right)-\frac 12$$ where $W(.)$ is Lambert function

The problem is that, with your numbers $(a=4,b=51.8682)$, the solution is a complex number.

Changing $b$ to its reciprocal would give $n=12.5120$ while its exact value is $12.5147$.