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CIE 9709 2023 March Paper 52 Q6

A Level / CIE / S1

CIE 9709 2023 March Paper 52 Paper · Question 6

题目

Problem

In a cycling event the times taken to complete a course are modelled by a normal distribution with mean 62.3 minutes and standard deviation 8.4 minutes.

(a) Find the probability that a randomly chosen cyclist has a time less than 74 minutes.

(2)

(b) Find the probability that 4 randomly chosen cyclists all have times between 50 and 74 minutes.

(4)

In a different cycling event, the times can also be modelled by a normal distribution. 23% of the cyclists have times less than 36 minutes and 10% of the cyclists have times greater than 54 minutes.

(c) Find estimates for the mean and standard deviation of this distribution.

(5)

题目中文翻译

在一项自行车赛事中，完成赛程所需的时间用均值为 62.3 分钟、标准差为 8.4 分钟的正态分布建模。

(a) 随机选择一名自行车运动员，求其用时少于 74 分钟的概率。

(2)

(b) 随机选择 4 名自行车运动员，求他们的用时全部在 50 至 74 分钟之间的概率。

(4)

在另一项自行车赛事中，用时也可用正态分布建模。23% 的运动员用时少于 36 分钟，10% 的运动员用时超过 54 分钟。

(c) 求该分布的均值和标准差的估计值。

(5)

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

标准化74分钟，求左尾概率。

答题过程

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Let T∼N(62.3,8.42)T\sim\mathrm{N}(62.3,8.4^2). Then

P(T<74)=Φ(74−62.38.4)=0.918168502…≈0.918.\mathrm{P}(T<74)=\Phi\biggl(\frac{74-62.3}{8.4}\biggr) =0.918168502\ldots\approx\boxed{0.918}.

(b)

解法一

思路

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先求一人的区间概率，再四次独立相乘，不要只把单人概率当最终答案。

答题过程

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For one cyclist,

p= P(50<T<74)= Φ(74−62.38.4)−Φ(50−62.38.4)= 0.846610553….\begin{align*} p=&\,\mathrm{P}(50<T<74)\\[4mm] =&\,\Phi\biggl(\frac{74-62.3}{8.4}\biggr) -\Phi\biggl(\frac{50-62.3}{8.4}\biggr)\\[4mm] =&\,0.846610553\ldots. \end{align*}

For four independently chosen cyclists,

P(all four)=p4=0.513729744…≈0.514.\mathrm{P}(\text{all four})=p^4=0.513729744\ldots\approx\boxed{0.514}.

(c)

解法一

思路

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第二个条件是右尾0.10，换为左尾0.90；两个分位点列联立方程。

答题过程

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For the other event, write T∼N(μ,σ2)T\sim\mathrm{N}(\mu,\sigma^2). Then

36−μσ=Φ−1(0.23)=−0.738846849…,\frac{36-\mu}{\sigma}=\Phi^{-1}(0.23)=-0.738846849\ldots, 54−μσ=Φ−1(0.90)=1.281551565….\frac{54-\mu}{\sigma}=\Phi^{-1}(0.90)=1.281551565\ldots.

Subtracting gives

σ= 181.281551565…+0.738846849…= 8.909133895….\begin{align*} \sigma=&\,\frac{18}{1.281551565\ldots+0.738846849\ldots}\\[4mm] =&\,8.909133895\ldots. \end{align*}

Then

μ= 36+0.738846849…(8.909133895…)= 42.582485507….\begin{align*} \mu=&\,36+0.738846849\ldots(8.909133895\ldots)\\[4mm] =&\,42.582485507\ldots. \end{align*}

Hence

μ≈42.6,σ≈8.91(minutes).\boxed{\mu\approx42.6,\qquad\sigma\approx8.91} \qquad\text{(minutes)}.