CalcGospel 國際數學圖譜
返回

CIE 9709 2025 November Paper 53 Q7

A Level / CIE / S1

CIE 9709 2025 Nov Paper 53 Paper · Question 7

题目

Problem

The times taken by Obi to walk to work each morning are normally distributed with mean 14.8 minutes and standard deviation 1.5 minutes.

(a) Find the probability that, on a randomly chosen day, Obi takes more than 15.6 minutes to walk to work.

(3)

(b) On 90% of days, Obi takes more than tt minutes to walk to work.

Find the value of tt.

(3)

Obi walks to work 5 days a week for 45 weeks in a year.

(c) On how many days in a year would you expect Obi to take within one minute of the mean time to walk to work?

(4)
题目中文翻译

Obi 每天早晨步行上班所用的时间服从正态分布，平均数为 14.8 分钟，标准差为 1.5 分钟。

(a) 求随机选取的一天中，Obi 步行上班所用时间多于 15.6 分钟的概率。

(3)

(b) 在 90% 的日子里，Obi 步行上班所用的时间多于 tt 分钟。

求 tt 的值。

(3)

Obi 每年有 45 周会步行上班，每周 5 天。

(c) 预计一年中有多少天，Obi 步行上班所用的时间与平均数相差不超过一分钟？

(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

将15.6标准化后取右尾，边界比均值大，所以所求概率应小于一半。

答题过程

展开

Let T∼N(14.8,1.52)T\sim\mathrm{N}(14.8,1.5^2) denote the time in minutes. Then

P(T>15.6)= 1−Φ(15.6−14.81.5)= 1−Φ(0.533333…)= 0.296901…≈ 0.297.\begin{align*} \mathrm{P}(T>15.6) =&\,1-\Phi\biggl(\frac{15.6-14.8}{1.5}\biggr)\\[4mm] =&\,1-\Phi(0.533333\ldots)\\[4mm] =&\,0.296901\ldots\\[4mm] \approx&\,\boxed{0.297}. \end{align*}

(b)

解法一

思路

展开

右侧90%等价于左侧10%，因此t低于均值。查左尾0.10的负标准分数，再还原时间。

答题过程

展开

Since P(T>t)=0.90\mathrm{P}(T>t)=0.90, we have P(T<t)=0.10\mathrm{P}(T<t)=0.10. Thus

t−14.81.5=Φ−1(0.10)=−1.281551…,\frac{t-14.8}{1.5}=\Phi^{-1}(0.10)=-1.281551\ldots,

so

t= 14.8+1.5(−1.281551…)= 12.877672…≈ 12.9.\begin{align*} t=&\,14.8+1.5(-1.281551\ldots)\\[4mm] =&\,12.877672\ldots\\[4mm] \approx&\,\boxed{12.9}. \end{align*}

(c)

解法一

思路

展开

均值一分以内的区间是13.8到15.8，总宽度两分钟。先求对称区间概率，再乘全年步行上班的225天。

答题过程

展开

The required interval is 13.8<T<15.813.8<T<15.8. Therefore,

P(∣T−14.8∣<1)= P(−11.5<Z<11.5)= 2Φ(2/3)−1= 0.495014….\begin{align*} \mathrm{P}(|T-14.8|<1) =&\,\mathrm{P}\biggl(-\frac1{1.5}<Z<\frac1{1.5}\biggr)\\[4mm] =&\,2\Phi(2/3)-1\\[4mm] =&\,0.495014\ldots. \end{align*}

Obi walks to work on 5×45=2255\times45=225 days. The expected number is

225(0.495014…)=111.378…,225(0.495014\ldots)=111.378\ldots,

giving 111 days.