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CIE 9709 2023 March Paper 62 Q5

A Level / CIE / S2

CIE 9709 2023 March Paper 62 Paper · Question 5

题目

Problem

The masses, in grams, of large and small packets of Maxwheat cereal have the independent distributions N(410.0,3.62)\mathrm{N}(410.0,3.6^2) and N(206.0,3.72)\mathrm{N}(206.0,3.7^2) respectively.

(a) Find the probability that a randomly chosen large packet has a mass that is more than double the mass of a randomly chosen small packet.

(5)

The packets are placed in boxes. The boxes are identical in appearance. 60%60\% of the boxes contain exactly 1010 randomly chosen large packets. 40%40\% of the boxes contain exactly 2020 randomly chosen small packets.

(b) Find the probability that a randomly chosen box contains packets with a total mass of more than 40804080 grams.

(6)
题目中文翻译

Maxwheat 谷物的大袋和小袋的质量（以克为单位）相互独立，分别服从 N(410.0,3.62)\mathrm{N}(410.0,3.6^2) 和 N(206.0,3.72)\mathrm{N}(206.0,3.7^2) 分布。

(a) 随机选取一大袋和一小袋，求大袋的质量超过小袋质量两倍的概率。

(5)

这些袋装谷物被装入箱中。各箱外观相同。其中 60%60\% 的箱子恰好装有随机选取的 1010 大袋谷物，40%40\% 的箱子恰好装有随机选取的 2020 小袋谷物。

(b) 随机选取一箱，求该箱内袋装谷物的总质量超过 40804080 克的概率。

(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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大包减去同一个小包的两倍为正态变量；小包方差须乘4。

答题过程

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Let LL and SS be the large- and small-packet masses. By independence,

D=L−2S.D=L-2S.

Then

D∼ N(410−2(206), 3.62+22(3.7)2)= N(−2,67.72).\begin{align*} D\sim&\,\mathrm{N}\bigl(410-2(206),\ 3.6^2+2^2(3.7)^2\bigr)\\[4mm] =&\,\mathrm{N}(-2,67.72). \end{align*}

Hence

P(L>2S)= 1−Φ(0−(−2)67.72)= 0.403989…= 0.404(3 s.f.).\begin{align*} \mathrm{P}(L>2S)=&\,1-\Phi\bigg(\frac{0-(-2)}{\sqrt{67.72}}\bigg)\\[4mm] =&\,0.403989\ldots\\[4mm] =&\,\boxed{0.404}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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两种盒装的总质量分布分别求出，先算各自超过4080的概率，再以0.6和0.4作全概率加权。

答题过程

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Let TLT_L and TST_S be the total packet masses for the two box types. Then

TL∼ N(10(410), 10(3.6)2)= N(4100,129.6),\begin{align*} T_L\sim&\,\mathrm{N}\bigl(10(410),\ 10(3.6)^2\bigr)\\[4mm] =&\,\mathrm{N}(4100,129.6), \end{align*}

and

TS∼ N(20(206), 20(3.7)2)= N(4120,273.8).\begin{align*} T_S\sim&\,\mathrm{N}\bigl(20(206),\ 20(3.7)^2\bigr)\\[4mm] =&\,\mathrm{N}(4120,273.8). \end{align*}

Thus

pL=1−Φ(4080−4100129.6)=0.9605258…,p_L=1-\Phi\bigg(\frac{4080-4100}{\sqrt{129.6}}\bigg) =0.9605258\ldots,

and

pS=1−Φ(4080−4120273.8)=0.9921835….p_S=1-\Phi\bigg(\frac{4080-4120}{\sqrt{273.8}}\bigg) =0.9921835\ldots.

By conditioning on the type of box,

 P(total mass>4080)= 0.6pL+0.4pS= 0.9731889…≈ 0.973(3 s.f.).\begin{align*} &\,\mathrm{P}(\text{total mass}>4080)\\[4mm] =&\,0.6p_L+0.4p_S\\[4mm] =&\,0.9731889\ldots\\[4mm] \approx&\,\boxed{0.973}\quad\text{(3 s.f.).} \end{align*}