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CIE 9709 2022 March Paper 52 Q4

A Level / CIE / S1

CIE 9709 2022 March Paper 52 Paper · Question 4

题目

Problem

The weights of male leopards in a particular region are normally distributed with mean 55 kg55\,\mathrm{kg} and standard deviation 6 kg6\,\mathrm{kg}.

(a) Find the probability that a randomly chosen male leopard from this region weighs between 46 and 62 kg62\,\mathrm{kg}.

(4)

The weights of female leopards in this region are normally distributed with mean 42 kg42\,\mathrm{kg} and standard deviation σ kg\sigma\,\mathrm{kg}. It is known that 25% of female leopards in the region weigh less than 36 kg36\,\mathrm{kg}.

(b) Find the value of σ\sigma.

(3)

The distributions of the weights of male and female leopards are independent of each other. A male leopard and a female leopard are each chosen at random.

(c) Find the probability that both the weights of these leopards are less than 46 kg46\,\mathrm{kg}.

(4)

题目中文翻译

某地区雄性豹的体重服从平均数为 55 kg55\,\mathrm{kg}、标准差为 6 kg6\,\mathrm{kg} 的正态分布。

(a) 求从该地区随机选取的一只雄性豹的体重在 46 至 62 kg62\,\mathrm{kg} 之间的概率。

(4)

该地区雌性豹的体重服从平均数为 42 kg42\,\mathrm{kg}、标准差为 σ kg\sigma\,\mathrm{kg} 的正态分布。已知该地区 25% 的雌性豹体重小于 36 kg36\,\mathrm{kg}。

(b) 求 σ\sigma 的值。

(3)

雄性豹和雌性豹的体重分布相互独立。分别随机选取一只雄性豹和一只雌性豹。

(c) 求这两只豹的体重都小于 46 kg46\,\mathrm{kg} 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

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雄豹标准差6，边界分别标准化为−1.5与7/6。

答题过程

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For a male leopard’s weight M∼N(55,62)M\sim\mathrm{N}(55,6^2),

P(46<M<62)= Φ(62−556) −Φ(46−556)= Φ(7/6)−Φ(−1.5)= 0.811520294…≈0.812.\begin{align*} \mathrm{P}(46<M<62)=&\,\Phi\biggl(\frac{62-55}{6}\biggr)\\[4mm] &\,\hspace{2pt}-\Phi\biggl(\frac{46-55}{6}\biggr)\\[4mm] =&\,\Phi(7/6)-\Phi(-1.5)\\[4mm] =&\,0.811520294\ldots\approx\boxed{0.812}. \end{align*}

(b)

解法一

思路

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36kg是雌豹第25百分位，对应标准正态负分位点。

答题过程

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For a female leopard’s weight F∼N(42,σ2)F\sim\mathrm{N}(42,\sigma^2),

36−42σ=Φ−1(0.25)=−0.674489750….\frac{36-42}{\sigma}=\Phi^{-1}(0.25)=-0.674489750\ldots.

Hence

σ=60.674489750…=8.895613311…≈8.90 kg.\sigma=\frac6{0.674489750\ldots}=8.895613311\ldots \approx\boxed{8.90\text{ kg}}.

(c)

解法一

思路

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分别求两只豹小于46的概率，由独立性相乘，沿用未舍入标准差。

答题过程

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For the male,

P(M<46)=Φ(−1.5)=0.066807201….\mathrm{P}(M<46)=\Phi(-1.5)=0.066807201\ldots.

For the female, using the standard deviation from (b),

P(F<46)=Φ(46−428.895613311…)=0.673522131….\mathrm{P}(F<46)=\Phi\biggl(\frac{46-42}{8.895613311\ldots}\biggr) =0.673522131\ldots.

By independence,

P(M<46, F<46)= (0.066807201…)(0.673522131…)= 0.044996129…≈0.0450.\begin{align*} \mathrm{P}(M<46,\,F<46)=&\,(0.066807201\ldots)(0.673522131\ldots)\\[4mm] =&\,0.044996129\ldots\approx\boxed{0.0450}. \end{align*}