5

A crabs' life expectancy can be modeled exponentially, and a crab lives 3 months on average.

I am absolutely not sure about this, because there is nothing concerning this in our book, so I guess it was meant to be solved in some obvious/easy fashion, here's what I tried:

If it were only one crab, I could simply plug 9 into $\lambda e^{-\lambda x}$, where $\lambda=1/3$.

$60$ is $10\%$ of $600$, so maybe I need to look after what time 90% died, intuitively I would resort to

$$1-e^{-x/3}=0.9$$ $$0.1=e^{-x/3}$$ and so on, which would give me $\approx 6.9$ months, and then do something about the remaining $2.1$ months.

The last thing I was thinking of is to calculate the probability for 540 crabs dying at some point before the 9 month mark, and then taking the converse probability, but that I'd only know how to do with the help of a computer.

Community
  • 1
355durch113
  • 1,570

3 Answers3

3

This is a two-step process. First, given the information about a crab's life expectancy, figure out the probability that an individual crab will be alive after 9 months. Let that value be $p$.

So now take the population of 600 crabs. Each of them has an independent (I assume) probability of being alive after 9 months of $p$. That means you've got a group of identical events - for each of the 600 crabs, it's either alive or dead, and the probability for each crab is the same value $p$. And we want to know something about the distribution of how many crabs are alive. What does that sound like?

ConMan
  • 24,300
1

The probability that a single crab with lifetime $L$ is alive after 9 months is $P[L > 9] = \int_{9}^{\infty}\frac{1}{3}\mathrm{e}^{-x/3} dx = \mathrm{e}^{-3}$.

The indicator of the event that $L > 9$ is Bernoulli with probability $\mathrm{e}^{-3}$.

The number of crabs, $N$, alive after 9 months will be a sum of these Bernoullis, i.e. $N$ is binomial (See here under the Bernoulli distribution subsection: https://en.wikipedia.org/wiki/Binomial_distribution#Related_distributions) with parameters 600 and $\mathrm{e}^{-3}$. So: $P[N = 60] = {{600}\choose{60}} \mathrm{e}^{-(3)(60)}(1 - \mathrm{e}^{-3})^{540}$.

Davis Yoshida
  • 1,530
1

For each crab, consider the CDF for the time of death: $$ F(t)=\Pr(\text{dying at or before time }t)=1-\exp(-\lambda t),\quad \lambda=\frac{1}{3}. $$ Assume that the 600 crabs are identical and do not affect one another. Let $X_i=1$ if crab $i$ is alive after 9 months and $0$ otherwise. Then, $\Pr(X_i=1)=p\equiv 1-F(9)$ and $\Pr(X_i=0)=1-p$. Then, $$ L\equiv\#\text{ crabs alive after 9 months}=\sum_{i=1}^{600}X_i $$ has binomial distribution $B(n,p)$ where $n=600$. You can use the usual formula then to compute $\Pr(L=60)$ or $\Pr(L\geq 60)$. Your question doesn't make clear which one you seek. If you seek the latter, an approximation is possible via a central limit theorem: $$ \frac{L}{\sqrt{n}}=\frac{1}{\sqrt{n}}\sum_{i=1}^{n}X_i\approx N[p,p(1-p)]. $$ You then can approximate \begin{align*} \Pr(L\geq 60)&=\Pr\left(\frac{\frac{L}{\sqrt{n}}-p}{\sqrt{p(1-p)}}\geq\frac{\frac{60}{\sqrt{n}}-p}{\sqrt{p(1-p)}}\right)\\ &\approx 1-\Phi\left(\frac{\frac{60}{\sqrt{n}}-p}{\sqrt{p(1-p)}}\right) \end{align*} where $\Phi(\cdot)$ denotes the CDF of $N(0,1)$.

yurnero
  • 10,505