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IAL 2024 Oct S2 Q1

A Level / Edexcel / S2

IAL 2024 Oct Paper · Question 1

题目

Problem

During an annual beach-clean, the people doing the clean are asked to conduct a litter survey. At a particular beach-clean, litter was found at a rate of 4 items per square metre.

(a) Find the probability that, in a randomly selected area of 2 square metres on this beach, exactly 5 items of litter were found.

(2)

Of the litter found on the beach, 30% of the items were face masks.

(b) Find the probability that, in a randomly selected area of 5 square metres on this beach, more than 4 face masks were found.

(2)

(c) Using a suitable approximation, find the probability that, in a randomly selected area of 20 square metres on this beach, less than 60 items of litter were found that were not face masks.

(4)

(Total for Question 1 is 8 marks)

题目中文翻译

在一次年度海滩清洁活动中，清洁人员被要求进行垃圾调查。在一次特定的海滩清洁中，垃圾的发现速率为每平方米 4 件。

(a) 求在此海滩上随机选择的 2 平方米区域内恰好发现 5 件垃圾的概率。

在海滩上发现的垃圾中，30% 为口罩。

(b) 求在此海滩上随机选择的 5 平方米区域内发现超过 4 个口罩的概率。

(c) 使用适当的近似方法，求在此海滩上随机选择的 20 平方米区域内发现少于 60 件非口罩垃圾的概率。

（第 1 题共 8 分）

解答

(a)

解法一

思路

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每平方米平均发现 4 件垃圾，因此 2 平方米内的平均数为 8。固定区域内的垃圾件数可用泊松分布建模，所求是恰好 5 件的概率。

答题过程

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Let XX be the number of litter items found in an area of 2 m22\text{ m}^2. Then

X∼Po⁡(8).X\sim\operatorname{Po}(8).

Therefore,

P(X=5)= e−8855!= 0.091603…≈ 0.0916.\begin{align*} P(X=5)=&\,\frac{e^{-8}8^5}{5!}\\ =&\,0.091603\ldots\\ \approx&\,\boxed{0.0916}. \end{align*}

(b)

解法一

思路

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5 平方米内的垃圾平均数为 4×5=204\times5=20，其中 30% 为口罩，所以口罩的平均数为 6。利用泊松分布的分类性质，口罩数服从均值为 6 的泊松分布；“超过 4 个”可用补事件计算。

答题过程

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Let YY be the number of face masks found in an area of 5 m25\text{ m}^2. Its mean is

4(5)(0.3)=6,4(5)(0.3)=6,

so

Y∼Po⁡(6).Y\sim\operatorname{Po}(6).

Therefore,

P(Y>4)= 1−P(Y≤4)= 1−0.2851= 0.7149≈ 0.715.\begin{align*} P(Y>4)=&\,1-P(Y\leq4)\\ =&\,1-0.2851\\ =&\,\boxed{0.7149}\\ \approx&\,\boxed{0.715}. \end{align*}

(c)

解法一

思路

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20 平方米内垃圾总数的平均数为 80，其中 70% 不是口罩，所以非口罩垃圾数服从均值为 56 的泊松分布。由于均值较大，可用均值和方差均为 56 的正态分布近似。“少于 60”即不超过 59，连续性修正后的边界为 59.5。

答题过程

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Let WW be the number of litter items that are not face masks in an area of 20 m220\text{ m}^2. Its mean is

4(20)(0.7)=56.4(20)(0.7)=56.

Thus W∼Po⁡(56)W\sim\operatorname{Po}(56). Let VV have the suitable normal approximation

V∼N(56,56).V\sim N(56,56).

Applying a continuity correction,

P(W<60)=P(W≤59)≈ P(V<59.5)= P(Z<59.5−5656)= P(Z<0.4677…)= 0.6800…≈ 0.680.\begin{align*} P(W<60)=P(W\leq59) \approx&\,P(V<59.5)\\ =&\,P\bigg(Z<\frac{59.5-56}{\sqrt{56}}\bigg)\\ =&\,P(Z<0.4677\ldots)\\ =&\,0.6800\ldots\\ \approx&\,\boxed{0.680}. \end{align*}