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CIE 9709 2026 March Paper 62 Q5

A Level / CIE / S2

CIE 9709 2026 March Paper 62 Paper · Question 5

题目

Problem

Anita has a dice with sides numbered 11, 22, 33, 44, 55, 66. The score on one throw is denoted by XX.

(a) Assuming that the dice is fair, show that Var⁡(X)=3512\operatorname{Var}(X)=\frac{35}{12}.

(2)

Anita suspects that the dice is not fair. She thinks that low scores are more likely than high scores. She intends to throw the dice 5050 times, and if the mean score is less than 33, she will conclude that the dice is not fair.

It is now given that the dice is not fair and E(X)=2.5\mathrm{E}(X)=2.5.

(b) Assuming that Var⁡(X)=3512\operatorname{Var}(X)=\frac{35}{12}, calculate the probability of a Type II error. You may assume that a continuity correction is not needed.

(3)

(c) State what is meant by a Type II error in this context.

(1)

(d) The sample mean score from Anita’s 5050 throws of the dice was 2.92.9.

Explain why it is not possible for Anita to make a Type II error.

(1)
题目中文翻译

Anita 有一枚骰子，各面分别标有 11、22、33、44、55、66。用 XX 表示掷一次所得的点数。

(a) 假设骰子是均匀的，证明 Var⁡(X)=3512\operatorname{Var}(X)=\frac{35}{12}。

(2)

Anita 怀疑骰子不均匀。她认为掷出较低点数比较高点数更容易。她打算掷骰子 5050 次。如果平均点数小于 33，她就断定骰子不均匀。

现已知骰子不均匀，且 E(X)=2.5\mathrm{E}(X)=2.5。

(b) 假设 Var⁡(X)=3512\operatorname{Var}(X)=\frac{35}{12}，计算发生第二类错误的概率。可以假设不需要进行连续性修正。

(3)

(c) 说明在此情境中第二类错误的含义。

(1)

(d) Anita 掷骰子 5050 次所得点数的样本均值为 2.92.9。

解释为什么 Anita 不可能犯第二类错误。

(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

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公平骰子的六个分数等可能，先求一阶矩与二阶矩，再作方差的差。

答题过程

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For a fair die, each score has probability 1/61/6. Hence

E(X)=1+2+3+4+5+66=72,\mathrm E(X)=\frac{1+2+3+4+5+6}{6}=\frac72,

and

E(X2)=12+22+32+42+52+626=916.\mathrm E(X^2)=\frac{1^2+2^2+3^2+4^2+5^2+6^2}{6}=\frac{91}{6}.

Therefore

Var⁡(X)= 916−(72)2= 182−14712=3512.\begin{align*} \operatorname{Var}(X)=&\,\frac{91}{6}-\bigg(\frac72\bigg)^2\\[4mm] =&\,\frac{182-147}{12}=\boxed{\frac{35}{12}}. \end{align*}

(b)

解法一

思路

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真实均值已经变成 2.52.5。第二类错误是在骰子确实不公平时未落入拒绝域，即样本均值至少为 33；按题目指示不作连续性修正。

答题过程

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When the die is unfair with E(X)=2.5\mathrm E(X)=2.5 and Var⁡(X)=35/12\operatorname{Var}(X)=35/12, the Central Limit Theorem gives

X‾≈N(2.5,35/1250).\overline X\approx\mathrm N\bigg(2.5,\frac{35/12}{50}\bigg).

The test rejects fairness when X‾<3\overline X<3, so a Type II error occurs when X‾≥3\overline X\ge3. Without a continuity correction, as instructed,

 P(Type II error)≈ 1−Φ(3−2.535/(12×50))= 1−Φ(2.070197…)≈ 0.0192.\begin{align*} &\,\mathrm P(\text{Type II error})\\[4mm] \approx&\,1-\Phi\bigg(\frac{3-2.5}{\sqrt{35/(12\times50)}}\bigg)\\[4mm] =&\,1-\Phi(2.070197\ldots)\\[4mm] \approx&\,\boxed{0.0192}. \end{align*}

(c)

解法一

思路

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第二类错误必须提到骰子实际上不公平，不能只说样本结果不显著。

答题过程

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A Type II error is concluding that the die is fair when it is actually unfair, with low scores more likely than high scores.

(d)

解法一

思路

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第二类错误只会发生在未拒绝原假设时；本次观测已经落入拒绝域。

答题过程

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The observed mean is 2.9<32.9<3, so Anita rejects the null hypothesis that the die is fair. Since a Type II error requires not rejecting the null hypothesis when it is false, this decision cannot be a Type II error.