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CIE 9709 2025 June Paper 52 Q7

A Level / CIE / S1

CIE 9709 2025 June Paper 52 Paper · Question 7

题目

Problem

Kestrels are birds whose adult wingspans are normally distributed with mean 74.8 cm74.8\,\mathrm{cm} and standard deviation 3.2 cm3.2\,\mathrm{cm}. A random sample of 120 adult kestrels is selected.

(a) How many of these 120 adult kestrels would you expect to have wingspan between 72.4 cm72.4\,\mathrm{cm} and 76.3 cm76.3\,\mathrm{cm}?

(4)

The masses of adult kestrels are normally distributed with mean μ kg\mu\,\mathrm{kg} and standard deviation σ kg\sigma\,\mathrm{kg}. It is known that 20%20\% of adult kestrels have mass greater than 0.202 kg0.202\,\mathrm{kg} and 28%28\% have mass less than 0.185 kg0.185\,\mathrm{kg}.

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

(5)

10 adult kestrels are selected at random.

(c) Find the probability that fewer than 3 have masses greater than 0.202 kg0.202\,\mathrm{kg}.

(3)
题目中文翻译

红隼是一种鸟，其成鸟的翼展服从正态分布，均值为 74.8 cm74.8\,\mathrm{cm}，标准差为 3.2 cm3.2\,\mathrm{cm}。随机选取 120 只成年红隼作为样本。

(a) 预计这 120 只成年红隼中有多少只的翼展在 72.4 cm72.4\,\mathrm{cm} 与 76.3 cm76.3\,\mathrm{cm} 之间？

(4)

成年红隼的质量服从正态分布，均值为 μ kg\mu\,\mathrm{kg}，标准差为 σ kg\sigma\,\mathrm{kg}。已知 20%20\% 的成年红隼质量大于 0.202 kg0.202\,\mathrm{kg}，28%28\% 的成年红隼质量小于 0.185 kg0.185\,\mathrm{kg}。

(b) 求 μ\mu 和 σ\sigma 的值。

(5)

随机选取 10 只成年红隼。

(c) 求其中质量大于 0.202 kg0.202\,\mathrm{kg} 的红隼少于 3 只的概率。

(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

将两个翼展边界标准化，求区间概率后乘120，原变量连续，不作连续性修正。

答题过程

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Let W∼N(74.8,3.22)W\sim\mathrm{N}(74.8,3.2^2) denote wingspan in centimetres. Then

P(72.4<W<76.3)= P(72.4−74.83.2<Z<76.3−74.83.2)= Φ(0.46875)−Φ(−0.75)= 0.453748….\begin{align*} \mathrm{P}(72.4<W<76.3) =&\,\mathrm{P}\biggl(\frac{72.4-74.8}{3.2}<Z <\frac{76.3-74.8}{3.2}\biggr)\\[4mm] =&\,\Phi(0.46875)-\Phi(-0.75)\\[4mm] =&\,0.453748\ldots. \end{align*}

The expected number is 120(0.453748…)=54.4498…120(0.453748\ldots)=54.4498\ldots, giving 54 kestrels.

(b)

解法一

思路

展开

0.202对应左尾0.80，0.185对应左尾0.28。两条标准化等式相减先求标准差，再代回求均值。

答题过程

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The standard normal quantiles are

Φ−1(0.80)=0.841621…,Φ−1(0.28)=−0.582841….\Phi^{-1}(0.80)=0.841621\ldots,\qquad \Phi^{-1}(0.28)=-0.582841\ldots.

Thus

0.202−μ= 0.841621… σ,0.185−μ= −0.582841… σ.\begin{align*} 0.202-\mu=&\,0.841621\ldots\,\sigma,\\[4mm] 0.185-\mu=&\,-0.582841\ldots\,\sigma. \end{align*}

Subtracting,

0.017= (0.841621…+0.582841…)σ,σ= 0.0119343….\begin{align*} 0.017=&\,(0.841621\ldots+0.582841\ldots)\sigma,\\[4mm] \sigma=&\,0.0119343\ldots. \end{align*}

Then μ=0.202−0.841621… σ=0.1919558…\mu=0.202-0.841621\ldots\,\sigma=0.1919558\ldots, so

μ≈0.192,σ≈0.0119.\boxed{\mu\approx0.192,\qquad\sigma\approx0.0119}.

(c)

解法一

思路

展开

成功概率由题设直接给出0.20，10只中满足条件的只数服从二项分布，少于3即0、1、2。

答题过程

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Let XX be the number with masses greater than 0.202 kg0.202\,\mathrm{kg}. Then X∼B(10,0.20)X\sim\mathrm{B}(10,0.20).

P(X<3)= (0.8)10+10(0.2)(0.8)9 +45(0.2)2(0.8)8= 0.6777995264≈ 0.678.\begin{align*} \mathrm{P}(X<3)=&\,(0.8)^{10}+10(0.2)(0.8)^9\\[4mm] &\,\hspace{2pt}+45(0.2)^2(0.8)^8\\[4mm] =&\,0.6777995264\\[4mm] \approx&\,\boxed{0.678}. \end{align*}