Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2022 November Paper 53 Q5

A Level / CIE / S1

CIE 9709 2022 Nov Paper 53 Paper · Question 5

题目

Problem

Company AA produces bags of sugar. An inspector finds that on average 10% of the bags are underweight.

10 of the bags are chosen at random.

(a) Find the probability that fewer than 3 of these bags are underweight.

(3)

The weights of the bags of sugar produced by company BB are normally distributed with mean 1.04 kg1.04\,\mathrm{kg} and standard deviation 0.06 kg0.06\,\mathrm{kg}.

(b) Find the probability that a randomly chosen bag produced by company BB weighs more than 1.11 kg1.11\,\mathrm{kg}.

(3)

81% of the bags of sugar produced by company BB weigh less than w kgw\,\mathrm{kg}.

(c) Find the value of ww.

(3)

题目中文翻译

公司 AA 生产袋装糖。一名检验员发现,平均有 10% 的袋装糖重量不足。

随机选取其中 10 袋。

(a) 求这些袋装糖中重量不足的少于 3 袋的概率。

(3)

公司 BB 生产的袋装糖的重量服从正态分布,平均数为 1.04 kg1.04\,\mathrm{kg},标准差为 0.06 kg0.06\,\mathrm{kg}。

(b) 求随机选取的一袋公司 BB 生产的糖的重量超过 1.11 kg1.11\,\mathrm{kg} 的概率。

(3)

公司 BB 生产的袋装糖中,81% 的重量小于 w kgw\,\mathrm{kg}。

(c) 求 ww 的值。

(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

重量不足袋数服从二项分布,少于3袋求0、1、2三项。

答题过程

展开

Let U∼B(10,0.1)U\sim\mathrm{B}(10,0.1). Then

P(U<3)= (0.9)10+10(0.1)(0.9)9 +(102)(0.1)2(0.9)8= 0.9298091736≈0.930.\begin{align*} \mathrm{P}(U<3)=&\,(0.9)^{10}+10(0.1)(0.9)^9\\[4mm] &\,\hspace{2pt}+\binom{10}{2}(0.1)^2(0.9)^8\\[4mm] =&\,0.9298091736\approx\boxed{0.930}. \end{align*}

(b)

解法一

思路

展开

公司B重量使用正态模型,标准化1.11并求右尾。

答题过程

展开

Let W∼N(1.04,0.062)W\sim\mathrm{N}(1.04,0.06^2). Then

P(W>1.11)= 1−Φ(1.11−1.040.06)= 0.121672504…≈0.122.\begin{align*} \mathrm{P}(W>1.11) =&\,1-\Phi\biggl(\frac{1.11-1.04}{0.06}\biggr)\\[4mm] =&\,0.121672504\ldots\approx\boxed{0.122}. \end{align*}

(c)

解法一

思路

展开

w对应左尾0.81,求正分位点后还原重量。

答题过程

展开 w−1.040.06=Φ−1(0.81)=0.877896295….\frac{w-1.04}{0.06}=\Phi^{-1}(0.81)=0.877896295\ldots.

Thus

w=1.04+0.06(0.877896295…)=1.092673777…≈1.09 kg.w=1.04+0.06(0.877896295\ldots) =1.092673777\ldots\approx\boxed{1.09\text{ kg}}.