Skip to content
CalcGospel 國際數學圖譜
返回

CIE 9709 2022 November Paper 61 Q3

A Level / CIE / S2

CIE 9709 2022 Nov Paper 61 Paper · Question 3

题目

Problem

Drops of water fall randomly from a leaking tap at a constant average rate of 5.25.2 per minute.

(a) Find the probability that at least 33 drops fall during a randomly chosen 3030-second period.

(3)

(b) Use a suitable approximating distribution to find the probability that at least 650650 drops fall during a randomly chosen 22-hour period.

(4)
题目中文翻译

水滴从一个漏水的水龙头随机滴落,平均速率恒为每分钟 5.25.2 滴。

(a) 随机选取一段 3030 秒的时间,求至少滴落 33 滴水的概率。

(3)

(b) 随机选取一段 22 小时的时间,用适当的近似分布求至少滴落 650650 滴水的概率。

(4)
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

30秒的泊松参数为2.6,至少3滴取不超过2滴的补集。

答题过程

展开

Let DD be the number of drops in 30 seconds. Then D∼Po(5.2/2)=Po(2.6)D\sim\mathrm{Po}(5.2/2)=\mathrm{Po}(2.6), so

P(D≥3)= 1−e−2.6(1+2.6+2.622!)≈ 0.482(3 s.f.).\begin{align*} \mathrm{P}(D\geq3)=&\,1-e^{-2.6}\bigg(1+2.6+\frac{2.6^2}{2!}\bigg)\\[4mm] \approx&\,\boxed{0.482}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

展开

两小时总数参数为624,用同均值方差正态近似,至少650的连续性修正边界为649.5。

答题过程

展开

For the two-hour total,

D120∼ Po(120(5.2))=Po(624)≈ N(624,624).\begin{align*} D_{120}\sim&\,\mathrm{Po}(120(5.2))=\mathrm{Po}(624)\\[4mm] \approx&\,\mathrm{N}(624,624). \end{align*}

Using a continuity correction,

P(D120≥650)≈ 1−Φ(649.5−624624)= 0.153671…= 0.154(3 s.f.).\begin{align*} \mathrm{P}(D_{120}\geq650)\approx&\,1-\Phi\bigg(\frac{649.5-624}{\sqrt{624}}\bigg)\\[4mm] =&\,0.153671\ldots\\[4mm] =&\,\boxed{0.154}\quad\text{(3 s.f.).} \end{align*}