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CIE 9709 2024 June Paper 51 Q2

A Level / CIE / S1

CIE 9709 2024 June Paper 51 Paper · Question 2

题目

Problem

The lengths of the tails of adult raccoons of a certain species are normally distributed with mean 28 cm28\,\mathrm{cm} and standard deviation 3.3 cm3.3\,\mathrm{cm}.

(a) Find the probability that a randomly chosen adult raccoon of this species has a tail length between 23 cm23\,\mathrm{cm} and 35 cm35\,\mathrm{cm}.

(4)

The masses of adult raccoons of this species are normally distributed with mean 8.5 kg8.5\,\mathrm{kg} and standard deviation σ kg\sigma\,\mathrm{kg}. 75%75\% of adult raccoons of this species have mass greater than 7.6 kg7.6\,\mathrm{kg}.

(b) Find the value of σ\sigma.

(3)
题目中文翻译

某种成年浣熊的尾长服从均值为 28 cm28\,\mathrm{cm}、标准差为 3.3 cm3.3\,\mathrm{cm} 的正态分布。

(a) 求随机选出的一只这种成年浣熊的尾长介于 23 cm23\,\mathrm{cm} 和 35 cm35\,\mathrm{cm} 之间的概率。

(4)

这种成年浣熊的质量服从均值为 8.5 kg8.5\,\mathrm{kg}、标准差为 σ kg\sigma\,\mathrm{kg} 的正态分布。这种成年浣熊中有 75%75\% 的质量大于 7.6 kg7.6\,\mathrm{kg}。

(b) 求 σ\sigma 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

将两个尾长边界标准化,求累计概率之差。

答题过程

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Let L∼N(28,3.32)L\sim\mathrm{N}(28,3.3^2). Then

P(23<L<35)= Φ(35−283.3)−Φ(23−283.3)= 0.918181…≈ 0.918.\begin{align*} \mathrm{P}(23<L<35) =&\,\Phi\biggl(\frac{35-28}{3.3}\biggr) -\Phi\biggl(\frac{23-28}{3.3}\biggr)\\[4mm] =&\,0.918181\ldots\\[4mm] \approx&\,\boxed{0.918}. \end{align*}

(b)

解法一

思路

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右尾0.75对应左尾0.25,其标准分数为负。

答题过程

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Let M∼N(8.5,σ2)M\sim\mathrm{N}(8.5,\sigma^2). Since P(M>7.6)=0.75\mathrm{P}(M>7.6)=0.75,

7.6−8.5σ=Φ−1(0.25)=−0.674489….\frac{7.6-8.5}{\sigma}=\Phi^{-1}(0.25)=-0.674489\ldots.

Therefore

σ=−0.9−0.674489…=1.334341…≈1.33.\sigma=\frac{-0.9}{-0.674489\ldots} =1.334341\ldots\approx\boxed{1.33}.