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IAL 2025 Jun S2 Q3

A Level / Edexcel / S2

IAL 2025 June Paper · Question 3

题目

Problem

A sweet shop produces different coloured sweets and mixes them to sell in small, medium or large bags.

The proportion of sweets that are blue is pp.

Each small bag is filled with a random sample of nn sweets.

The mean number of blue sweets in a small bag is 3 and the variance of the number of blue sweets in a small bag is 2.55

(a) Find the value of nn and the value of pp

(3)

The proportion of sweets that are green is 0.2

Each medium bag contains a random sample of 40 sweets.

For a medium bag,

(b) find the probability that the number of green sweets exceeds the expected number of green sweets.

(3)

The proportion of sweets that are yellow is 0.1

Each large bag contains a random sample of 80 sweets.

(c) (i) Use a Poisson approximation to find the probability that a large bag of sweets contains at least 12 yellow sweets.

(ii) Justify why a Poisson approximation is suitable in this case.

(4)

(Total for Question 3 is 10 marks)

题目中文翻译

一家糖果店生产不同颜色的糖果，并将它们混合装入小、中或大袋中出售。

蓝色糖果的比例为 pp。

每小袋装入 nn 颗糖果的随机样本。

一小袋中蓝色糖果的平均数为 3，方差为 2.55。

(a) 求 nn 的值和 pp 的值。

绿色糖果的比例为 0.2。

每中袋装入 40 颗糖果的随机样本。

对于一中袋

(b) 求绿色糖果数量超过期望数量的概率。

黄色糖果的比例为 0.1。

每大袋装入 80 颗糖果的随机样本。

(c) (i) 使用泊松近似求一大袋糖果中至少有 12 颗黄色糖果的概率。

(ii) 说明为什么泊松近似在此情况下是合适的。

（第 3 题共 10 分）

解答

(a)

解法一

思路

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小袋中蓝色糖果数服从二项分布，因此均值为 npnp，方差为 np(1−p)np(1-p)。把题目给出的 3 和 2.55 分别代入；用方差方程除以均值方程可先求 pp，再反求 nn。

答题过程

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Let BB be the number of blue sweets in a small bag. Then

B∼B⁡(n,p).B\sim\operatorname{B}(n,p).

Using the given mean and variance,

np=3np=3

and

np(1−p)=2.55.np(1-p)=2.55.

Dividing the second equation by the first gives

1−p=2.553=0.85,1-p=\frac{2.55}{3}=0.85,

so

p=0.15.p=0.15.

Therefore,

n=30.15=20.n=\frac3{0.15}=20.

Hence,

n=20,p=0.15.\boxed{n=20, \qquad p=0.15}.

(b)

解法一

思路

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中袋中绿色糖果数服从 B(40,0.2)B(40,0.2)，期望为 40(0.2)=840(0.2)=8。“超过期望数量”表示严格大于 8，即至少 9 颗；使用补事件计算 1−P(G≤8)1-P(G\leq8)。

答题过程

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Let GG be the number of green sweets in a medium bag. Then

G∼B⁡(40,0.2)G\sim\operatorname{B}(40,0.2)

and

E(G)=40(0.2)=8.E(G)=40(0.2)=8.

Therefore,

P(G>E(G))= P(G>8)= 1−P(G≤8)= 0.40687…= 0.407(3 s.f.).\begin{align*} P(G>E(G)) =&\,P(G>8)\\ =&\,1-P(G\leq8)\\ =&\,0.40687\ldots\\ =&\,\boxed{0.407}\quad\text{(3 s.f.)}. \end{align*}

(c)(i)

解法一

思路

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大袋中黄色糖果数原本服从 B(80,0.1)B(80,0.1)。泊松近似的均值为 np=8np=8；“至少 12 颗”表示 Y≥12Y\geq12，用补事件写成 1−P(Y≤11)1-P(Y\leq11)。

答题过程

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Let YY be the number of yellow sweets in a large bag. Using a Poisson approximation,

Y∼˙Po⁡(80×0.1)=Po⁡(8).Y\mathrel{\dot\sim}\operatorname{Po}(80\times0.1) =\operatorname{Po}(8).

Hence,

P(Y≥12)= 1−P(Y≤11)= 0.11192…= 0.112(3 s.f.).\begin{align*} P(Y\geq12) =&\,1-P(Y\leq11)\\ =&\,0.11192\ldots\\ =&\,\boxed{0.112}\quad\text{(3 s.f.)}. \end{align*}

(c)(ii)

解法一

思路

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二项分布可用泊松分布近似的典型条件是试验次数较大而单次成功概率较小。本题 n=80n=80 较大，p=0.1p=0.1 较小，符合条件。

答题过程

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The Poisson approximation is suitable because

n=80 is large and p=0.1 is small.\boxed{n=80\text{ is large and }p=0.1\text{ is small}}.