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CIE 9709 2026 June Paper 65 Q1

A Level / CIE / S2

CIE 9709 2026 June Paper 65 Paper · Question 1

题目

Problem

It is known that, on average, 11 in 500500 plants of a certain variety have a particular disease. A random sample of 250250 plants of this variety is selected, and the number of plants with the disease is denoted by XX.

(a) State the exact distribution of XX.

(1)

(b) State an appropriate approximating distribution for XX.

(1)

(c) Use your approximating distribution to find the probability that more than 22 plants in the sample have the disease.

(2)

Diseased plants occur at random and independently. Each day a scientist examines a selection of plants, and the number of diseased plants found per day is denoted by YY. The scientist finds that the mean of YY has approximately the same value every day, and this value is 3232.

(d) Use an appropriate approximating distribution to find P(Y<20)\mathrm{P}(Y<20).

(4)
题目中文翻译

已知某品种的植物平均每 500500 株中有 11 株患有某种病害。从该品种中随机抽取 250250 株植物,用 XX 表示其中患病的株数。

(a) 写出 XX 的精确分布。

(1)

(b) 写出 XX 的一个合适的近似分布。

(1)

(c) 用你选取的近似分布,求样本中患病植物超过 22 株的概率。

(2)

患病植物随机且相互独立地出现。一名科学家每天检查一些植物,用 YY 表示每天发现的患病株数。科学家发现 YY 的均值每天大致相同,均为 3232。

(d) 用一个合适的近似分布求 P(Y<20)\mathrm{P}(Y<20)。

(4)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

固定抽取 250250 株,每株是否患病是二项试验,成功概率为 1/5001/500。

答题过程

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There are 250250 randomly selected plants, each with disease probability 1/5001/500. Thus

X∼B(250,1500).\boxed{X\sim\mathrm B\bigg(250,\frac1{500}\bigg)}.

(b)

解法一

思路

展开

二项分布中试验次数大、成功概率小,使用参数为 npnp 的 Poisson 近似。

答题过程

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Since 250250 is large and 1/5001/500 is small, an appropriate approximation is

X≈Po(250×1500)=Po(0.5).\boxed{X\approx\mathrm{Po}\bigg(250\times\frac1{500}\bigg)=\mathrm{Po}(0.5)}.

(c)

解法一

思路

展开

“多于两株”是至少三株,使用补事件排除 0,1,20,1,2 三项。

答题过程

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Using the Poisson approximation,

P(X>2)≈ 1−e−0.5(1+0.5+0.522!)= 1−e−0.5(1.625)≈ 0.0144.\begin{align*} \mathrm P(X>2) \approx&\,1-\mathrm e^{-0.5}\bigg(1+0.5+\frac{0.5^2}{2!}\bigg)\\[4mm] =&\,1-\mathrm e^{-0.5}(1.625)\\[4mm] \approx&\,\boxed{0.0144}. \end{align*}

(d)

解法一

思路

展开

随机独立且平均数稳定,对每天患病株数建立 Poisson 模型。参数 3232 较大,再用正态近似并进行连续性修正。

答题过程

展开

Model YY by Po(32)\mathrm{Po}(32). Since 32>1532>15, use the normal approximation N∼N(32,32)N\sim\mathrm N(32,32).

With a continuity correction,

P(Y<20)= P(Y≤19)≈ P(N<19.5)= Φ(19.5−3232)= Φ(−2.20971…)≈ 0.0136.\begin{align*} \mathrm P(Y<20) =&\,\mathrm P(Y\le19)\\[4mm] \approx&\,\mathrm P(N<19.5)\\[4mm] =&\,\Phi\bigg(\frac{19.5-32}{\sqrt{32}}\bigg)\\[4mm] =&\,\Phi(-2.20971\ldots)\\[4mm] \approx&\,\boxed{0.0136}. \end{align*}