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CIE 9709 2022 November Paper 62 Q6

A Level / CIE / S2

CIE 9709 2022 Nov Paper 62 Paper · Question 6

题目

Problem

The masses, in grams, of small and large bags of flour have the distributions N(510,100)\mathrm{N}(510,100) and N(1015,324)\mathrm{N}(1015,324) respectively. André selects 44 small bags of flour and 22 large bags of flour at random.

(a) Find the probability that the total mass of these 66 bags of flour is less than 4130 g4130\,\mathrm{g}.

(5)

(b) Find the probability that the total mass of the 44 small bags is more than the total mass of the 22 large bags.

(5)
题目中文翻译

小袋面粉和大袋面粉的质量（以克为单位）分别服从 N(510,100)\mathrm{N}(510,100) 和 N(1015,324)\mathrm{N}(1015,324) 分布。André 随机选取 44 小袋和 22 大袋面粉。

(a) 求这 66 袋面粉的总质量小于 4130 g4130\,\mathrm{g} 的概率。

(5)

(b) 求 44 小袋的总质量大于 22 大袋的总质量的概率。

(5)
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

按随机选取各袋的独立抽样模型，四小袋与两大袋的均值方差分别相加，建立总质量的正态分布。

答题过程

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Let SS and LL be the total masses of the four small bags and the two large bags respectively. For the randomly selected bags,

S∼N(4(510),4(100))=N(2040,400),S\sim\mathrm{N}(4(510),4(100))=\mathrm{N}(2040,400),

and

L∼N(2(1015),2(324))=N(2030,648).L\sim\mathrm{N}(2(1015),2(324))=\mathrm{N}(2030,648).

Adding the independent masses gives

T=S+L.T=S+L.

Then

T∼N(4070,1048).T\sim\mathrm{N}(4070,1048).

Therefore

P(T<4130)= Φ(4130−40701048)≈ 0.968(3 s.f.).\begin{align*} \mathrm{P}(T<4130)=&\,\Phi\bigg(\frac{4130-4070}{\sqrt{1048}}\bigg)\\[4mm] \approx&\,\boxed{0.968}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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用前问两个独立总质量的差，均值相减、方差相加。

答题过程

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Using the totals from part (a),

D=S−L.D=S-L.

Then

D∼ N(2040−2030, 400+648)= N(10,1048).\begin{align*} D\sim&\,\mathrm{N}(2040-2030,\ 400+648)\\[4mm] =&\,\mathrm{N}(10,1048). \end{align*}

Hence

P(S>L)= 1−Φ(0−101048)= 0.621302…= 0.621(3 s.f.).\begin{align*} \mathrm{P}(S>L)=&\,1-\Phi\bigg(\frac{0-10}{\sqrt{1048}}\bigg)\\[4mm] =&\,0.621302\ldots\\[4mm] =&\,\boxed{0.621}\quad\text{(3 s.f.).} \end{align*}