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CIE 9709 2025 June Paper 61 Q2

A Level / CIE / S2

CIE 9709 2025 June Paper 61 Paper · Question 2

题目

Problem

The random variable XX has the distribution B(8,34)\mathrm{B}\big(8,\frac34\big). A random sample of 100100 values of XX is chosen, and the sample mean, X‾\overline{X}, is found.

(a) Find P(X‾>6.2)\mathrm{P}(\overline{X}>6.2). You are not expected to use a continuity correction.

(6)

(b) State why the Central Limit Theorem was needed in the calculation in part (a).

(1)
题目中文翻译

随机变量 XX 服从分布 B(8,34)\mathrm{B}\big(8,\frac34\big)。随机选取 XX 的 100100 个取值组成样本,并求出样本均值 X‾\overline{X}。

(a) 求 P(X‾>6.2)\mathrm{P}(\overline{X}>6.2)。不要求使用连续性修正。

(6)

(b) 说明为什么在 (a) 的计算中需要使用中心极限定理。

(1)
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

先求单个二项变量的均值与方差,再转为样本均值的均值和方差;按题意不做连续性修正。

答题过程

展开

For X∼B(8,3/4)X\sim\mathrm B(8,3/4),

E(X)= 8(34)=6,Var⁡(X)= 8(34)(14)=32.\begin{align*} \mathrm E(X)=&\,8\bigg(\frac34\bigg)=6,\\[4mm] \operatorname{Var}(X)=&\,8\bigg(\frac34\bigg)\bigg(\frac14\bigg)=\frac32. \end{align*}

By the Central Limit Theorem, for a random sample of size 100100,

X‾≈N(6,1.5100).\overline X\approx\mathrm N\bigg(6,\frac{1.5}{100}\bigg).

Without a continuity correction, as instructed,

P(X‾>6.2)≈ 1−Φ(6.2−61.5/100)= 1−Φ(1.632993…)≈ 0.0512.\begin{align*} \mathrm P(\overline X>6.2) \approx&\,1-\Phi\bigg(\frac{6.2-6}{\sqrt{1.5/100}}\bigg)\\[4mm] =&\,1-\Phi(1.632993\ldots)\\[4mm] \approx&\,\boxed{0.0512}. \end{align*}

(b)

解法一

思路

展开

原变量是二项分布而非正态;用中心极限定理来支持样本均值的正态近似。

答题过程

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The population distribution of XX is binomial, not normal. The Central Limit Theorem is needed to justify approximating the sample-mean distribution by a normal distribution for the large sample.