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CIE 9709 2022 November Paper 52 Q2

A Level / CIE / S1

CIE 9709 2022 Nov Paper 52 Paper · Question 2

题目

Problem

The lengths of the rods produced by a company are normally distributed with mean 55.6 mm55.6\,\mathrm{mm} and standard deviation 1.2 mm1.2\,\mathrm{mm}.

(a) In a random sample of 400 of these rods, how many would you expect to have length less than 54.8 mm54.8\,\mathrm{mm}?

(4)

(b) Find the probability that a randomly chosen rod produced by this company has a length that is within half a standard deviation of the mean.

(3)

题目中文翻译

某公司生产的杆的长度服从正态分布，均值为 55.6 mm55.6\,\mathrm{mm}，标准差为 1.2 mm1.2\,\mathrm{mm}。

(a) 从这些杆中随机抽取 400 根作为样本，预计其中有多少根的长度小于 54.8 mm54.8\,\mathrm{mm}？

(4)

(b) 随机选择该公司生产的一根杆，求其长度与均值相差不超过半个标准差的概率。

(3)

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

先算长度低于54.8的正态概率，再乘400，保留精度后取整数根数。

答题过程

展开

Let L∼N(55.6,1.22)L\sim\mathrm{N}(55.6,1.2^2). Then

P(L<54.8)=Φ(54.8−55.61.2)=0.252492537….\mathrm{P}(L<54.8)=\Phi\biggl(\frac{54.8-55.6}{1.2}\biggr) =0.252492537\ldots.

The expected number is

400(0.252492537…)=100.997015….400(0.252492537\ldots)=100.997015\ldots.

Thus we would expect 101 rods.

(b)

解法一

思路

展开

距均值半个标准差对应标准正态中央区间−0.5至0.5。

答题过程

展开 P(∣L−55.6∣<0.5(1.2))= P(−0.5<Z<0.5)= 2Φ(0.5)−1= 0.382924922…≈0.383.\begin{align*} \mathrm{P}(|L-55.6|<0.5(1.2)) =&\,\mathrm{P}(-0.5<Z<0.5)\\[4mm] =&\,2\Phi(0.5)-1\\[4mm] =&\,0.382924922\ldots\approx\boxed{0.383}. \end{align*}