Question

一位生物学家想估算湖中的鱼数。她在五月 $1$ 日随机捕获 $60$ 条鱼，给它们做上标记后放回湖中。在九月 $1$ 日，她随机捕获 $70$ 条鱼，发现其中 $3$ 条有标记。为估算五月 $1$ 日湖中的鱼数，她假设到九月 $1$ 日时这些鱼中有 $25\%$ 已不在湖中（因为死亡或迁出），湖中有 $40\%$ 的鱼在五月 $1$ 日时尚不在湖中（因为出生或迁入），并且九月 $1$ 日样本中有标记与无标记的鱼数能够代表总体。根据她的计算，五月 $1$ 日湖中有多少条鱼？

A biologist wants to calculate the number of fish in a lake. On May $1$ she catches a random sample of $60$ fish, tags them, and releases them. On September $1$ she catches a random sample of $70$ fish and finds that $3$ of them are tagged. To calculate the number of fish in the lake on May $1,$ she assumes that $25\%$ of these fish are no longer in the lake on September $1$ (because of death and emigrations), that $40\%$ of the fish were not in the lake May $1$ (because of births and immigrations), and that the number of untagged fish and tagged fish in the September $1$ sample are representative of the total population. What does the biologist calculate for the number of fish in the lake on May $1?$

答案

答案：840

知识点：百分数比与比例无放回抽样

难度评级：1830

小提示：

先求原先有标记的鱼中有多少条在九月仍留在湖中

First determine how many of the original tagged fish remain in September

大提示：

用样本中有标记的比例估算九月的鱼数，再确定其中五月时已经存在的 $60\%$

Use the sample’s tagged fraction to estimate the September population, then identify the $60\%$ that were present in May

解答：

在 $60$ 条有标记的鱼中， $75\%$ 到九月时仍在湖中，所以还剩 $45$ 条有标记的鱼。由样本估计，有标记的鱼占九月鱼群的 $\frac{3}{70}$ ，因此九月鱼群总数为 $45\left(\frac{70}{3}\right)=1050\text{。}$ 其中有 $60\%$ 在五月时已经存在，所以五月鱼群中仍存活的有 $630$ 条。这些鱼占五月鱼群的 $75\%$ 。因此估算出的五月鱼群总数为 $\frac{630}{0.75}=840$ 。

Of the $60$ tagged fish, $75\%$ remain in September, so $45$ tagged fish remain. The sample estimates that tagged fish form $\frac{3}{70}$ of the September population, making that population $45\left(\frac{70}{3}\right)=1050.$ Of those fish, $60\%$ were present in May, so $630$ surviving May fish remain. These are $75\%$ of the May population. Thus the estimated May population is $\frac{630}{0.75}=840.$

1990 AIME 第 6 题

答案：840

难度评级：1830

小提示：

大提示：

解答：

其他年份的第 6 题