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CIE 9709 2023 June Paper 51 Q4

A Level / CIE / S1

CIE 9709 2023 June Paper 51 Paper · Question 4

题目

Problem

A mathematical puzzle is given to a large number of students. The times taken to complete the puzzle are normally distributed with mean 14.6 minutes and standard deviation 5.2 minutes.

(a) In a random sample of 250 of the students, how many would you expect to have taken more than 20 minutes to complete the puzzle?

(4)

All the students are given a second puzzle to complete. Their times, in minutes, are normally distributed with mean μ\mu and standard deviation σ\sigma. It is found that 20% of the students have times less than 14.5 minutes and 67% of the students have times greater than 18.5 minutes.

(b) Find the value of μ\mu and the value of σ\sigma.

(5)

题目中文翻译

让大量学生完成一道数学谜题。他们完成这道谜题所需的时间服从正态分布，均值为 14.6 分钟，标准差为 5.2 分钟。

(a) 从这些学生中随机抽取 250 人，预计其中有多少人完成这道谜题的时间超过 20 分钟？

(4)

让所有学生完成第二道谜题。他们所需的时间（单位为分钟）服从均值为 μ\mu、标准差为 σ\sigma 的正态分布。发现 20% 的学生用时少于 14.5 分钟，67% 的学生用时超过 18.5 分钟。

(b) 求 μ\mu 和 σ\sigma 的值。

(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

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先求超过20分钟的概率，保留精度再乘250，最后给整数人数。

答题过程

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Let T∼N(14.6,5.22)T\sim\mathrm{N}(14.6,5.2^2). Then

P(T>20)= 1−Φ(20−14.65.2)= 0.149527616….\begin{align*} \mathrm{P}(T>20) =&\,1-\Phi\biggl(\frac{20-14.6}{5.2}\biggr)\\[4mm] =&\,0.149527616\ldots. \end{align*}

The expected number is

250(0.149527616…)=37.381904….250(0.149527616\ldots)=37.381904\ldots.

Thus we would expect 37 students.

(b)

解法一

思路

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右尾0.67换成左尾0.33，两个已知边界都低于均值，对应两个负分位点。

答题过程

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For the second puzzle,

14.5−μσ=Φ−1(0.20)=−0.841621233…,\frac{14.5-\mu}{\sigma}=\Phi^{-1}(0.20)=-0.841621233\ldots, 18.5−μσ=Φ−1(0.33)=−0.439913165….\frac{18.5-\mu}{\sigma}=\Phi^{-1}(0.33)=-0.439913165\ldots.

Subtracting these equations,

σ= 40.841621233…−0.439913165…= 9.957479870….\begin{align*} \sigma=&\,\frac4{0.841621233\ldots-0.439913165\ldots}\\[4mm] =&\,9.957479870\ldots. \end{align*}

Then

μ= 14.5+0.841621233…(9.957479870…)= 22.880426492….\begin{align*} \mu=&\,14.5+0.841621233\ldots(9.957479870\ldots)\\[4mm] =&\,22.880426492\ldots. \end{align*}

Thus

μ≈22.9,σ≈9.96(minutes).\boxed{\mu\approx22.9,\qquad\sigma\approx9.96} \qquad\text{(minutes)}.