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CIE 9709 2022 November Paper 51 Q4

A Level / CIE / S1

CIE 9709 2022 Nov Paper 51 Paper · Question 4

题目

Problem

In a large population, the systolic blood pressure (SBP) of adults is normally distributed with mean 125.4 and standard deviation 18.6.

(a) Find the probability that the SBP of a randomly chosen adult is less than 132.

(2)

The SBP of 12-year-old children in the same population is normally distributed with mean 117. Of these children 88% have SBP more than 108.

(b) Find the standard deviation of this distribution.

(3)

Three adults are chosen at random from this population.

(c) Find the probability that each of these three adults has SBP within 1.5 standard deviations of the mean.

(4)

题目中文翻译

在一个大型人群中,成年人的收缩压(SBP)服从正态分布,均值为 125.4,标准差为 18.6。

(a) 随机选择一名成年人,求其 SBP 小于 132 的概率。

(2)

同一人群中 12 岁儿童的 SBP 服从均值为 117 的正态分布。这些儿童中有 88% 的 SBP 大于 108。

(b) 求该分布的标准差。

(3)

从该人群中随机选择三名成年人。

(c) 求这三名成年人每人的 SBP 与均值相差均不超过 1.5 个标准差的概率。

(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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标准化132,求左尾概率。

答题过程

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For adult SBP S∼N(125.4,18.62)S\sim\mathrm{N}(125.4,18.6^2),

P(S<132)=Φ(132−125.418.6)=0.638644786…≈0.639.\mathrm{P}(S<132)=\Phi\biggl(\frac{132-125.4}{18.6}\biggr) =0.638644786\ldots\approx\boxed{0.639}.

(b)

解法一

思路

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88%高于108换成12%低于108,对应负分位点。

答题过程

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For the children, P(S<108)=0.12\mathrm{P}(S<108)=0.12. Thus

108−117σ=Φ−1(0.12)=−1.174986792…,\frac{108-117}{\sigma}=\Phi^{-1}(0.12)=-1.174986792\ldots,

so

σ=−9−1.174986792…=7.659660568…≈7.66.\sigma=\frac{-9}{-1.174986792\ldots} =7.659660568\ldots\approx\boxed{7.66}.

(c)

解法一

思路

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每人的边界已按标准差倍数给出,单人概率为标准正态中央区间概率,再三次独立相乘。

答题过程

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For one adult,

p= P(−1.5<Z<1.5)= 2Φ(1.5)−1=0.866385597….\begin{align*} p=&\,\mathrm{P}(-1.5<Z<1.5)\\[4mm] =&\,2\Phi(1.5)-1=0.866385597\ldots. \end{align*}

The probability that all three adults satisfy the condition is

p3=0.650329825…≈0.650.p^3=0.650329825\ldots\approx\boxed{0.650}.