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CIE 9709 2024 November Paper 62 Q5

A Level / CIE / S2

CIE 9709 2024 Nov Paper 62 Paper · Question 5

题目

Problem

A machine puts sweets into bags at random. The numbers of lemon and orange sweets in a bag have the independent distributions Po(3.7)\mathrm{Po}(3.7) and Po(2.6)\mathrm{Po}(2.6) respectively.

A bag of sweets is chosen at random.

(a) Find the probability that the number of lemon sweets in the bag is more than 22 but not more than 55.

(2)

(b) Find the probability that the total number of lemon and orange sweets in the bag is less than 44.

(3)

1010 bags of sweets are chosen at random.

(c) Use approximating distributions to find the probability that the total number of lemon sweets in the 1010 bags is less than the total number of orange sweets in the 1010 bags.

(6)
题目中文翻译

一台机器将糖果随机装入袋中。一袋中柠檬味糖果和橙味糖果的颗数相互独立，分别服从 Po(3.7)\mathrm{Po}(3.7) 和 Po(2.6)\mathrm{Po}(2.6) 分布。

随机选取一袋糖果。

(a) 求该袋中柠檬味糖果多于 22 颗且不多于 55 颗的概率。

(2)

(b) 求该袋中柠檬味和橙味糖果的总颗数少于 44 的概率。

(3)

随机选取 1010 袋糖果。

(c) 用近似分布求这 1010 袋中柠檬味糖果总颗数少于橙味糖果总颗数的概率。

(6)
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

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柠檬糖的个数须为3、4或5，直接累加三个泊松概率。

答题过程

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Let LL be the number of lemon sweets in one bag. Since L∼Po(3.7)L\sim\mathrm{Po}(3.7),

P(2<L≤5)= e−3.7(3.733!+3.744!) +e−3.73.755!= 0.544655…= 0.545(3 s.f.).\begin{align*} \mathrm{P}(2<L\leq5)=&\,e^{-3.7}\bigg(\frac{3.7^3}{3!}+\frac{3.7^4}{4!}\bigg)\\[4mm] &\,+e^{-3.7}\frac{3.7^5}{5!}\\[4mm] =&\,0.544655\ldots\\[4mm] =&\,\boxed{0.545}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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独立泊松变量之和仍为泊松分布，参数为3.7加2.6；总数小于4即不超过3。

答题过程

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Let SS be the total number of lemon and orange sweets in a bag. By independence,

S∼Po(3.7+2.6)=Po(6.3).S\sim\mathrm{Po}(3.7+2.6)=\mathrm{Po}(6.3).

Therefore

P(S<4)= e−6.3(1+6.3) +e−6.3(6.322!+6.333!)= 0.126374…= 0.126(3 s.f.).\begin{align*} \mathrm{P}(S<4)=&\,e^{-6.3}(1+6.3)\\[4mm] &\,+e^{-6.3}\bigg(\frac{6.3^2}{2!}+\frac{6.3^3}{3!}\bigg)\\[4mm] =&\,0.126374\ldots\\[4mm] =&\,\boxed{0.126}\quad\text{(3 s.f.).} \end{align*}

(c)

解法一

思路

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10袋的两种糖总数分别近似为正态变量。差的方差相加，严格少于对应整数差至少1，使用0.5的连续性修正边界。

答题过程

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Let L10L_{10} and O10O_{10} denote the totals in the 10 bags. Their distributions and normal approximations are

L10∼ Po(37)≈N(37,37),O10∼ Po(26)≈N(26,26).\begin{align*} L_{10}\sim&\,\mathrm{Po}(37)\approx\mathrm{N}(37,37),\\[4mm] O_{10}\sim&\,\mathrm{Po}(26)\approx\mathrm{N}(26,26). \end{align*}

By independence, the difference D=O10−L10D=O_{10}-L_{10} is approximated by

D≈ N(26−37, 26+37)= N(−11,63).\begin{align*} D\approx&\,\mathrm{N}(26-37,\ 26+37)\\[4mm] =&\,\mathrm{N}(-11,63). \end{align*}

The event L10<O10L_{10}<O_{10} is D≥1D\geq1. Using a continuity correction,

 P(L10<O10)≈ 1−Φ(0.5−(−11)63)= 0.0736878…= 0.0737(3 s.f.).\begin{align*} &\,\mathrm{P}(L_{10}<O_{10})\\[4mm] \approx&\,1-\Phi\bigg(\frac{0.5-(-11)}{\sqrt{63}}\bigg)\\[4mm] =&\,0.0736878\ldots\\[4mm] =&\,\boxed{0.0737}\quad\text{(3 s.f.).} \end{align*}