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CIE 9709 2022 June Paper 61 Q2

A Level / CIE / S2

CIE 9709 2022 June Paper 61 Paper · Question 2

题目

Problem

Arvind uses an ordinary fair 66-sided die to play a game. He believes he has a system to predict the score when the die is thrown. Before each throw of the die, he writes down what he thinks the score will be. He claims that he can write the correct score more often than he would if he were just guessing.

His friend Laxmi tests his claim by asking him to write down the score before each of 1515 throws of the die. Arvind writes the correct score on exactly 55 out of 1515 throws.

Test Arvind’s claim at the 10%10\% significance level.

(5)
题目中文翻译

Arvind 用一枚普通的公平六面骰子玩游戏。他认为自己有一套方法能预测掷骰所得的点数。每次掷骰前，他都写下自己预测的点数。他声称，自己写对点数的次数比单靠猜测时更多。

他的朋友 Laxmi 让他在 1515 次掷骰前逐次写下预测点数，以检验他的说法。这 1515 次中，Arvind 恰好有 55 次写对了点数。

在 10%10\% 的显著性水平下检验 Arvind 的说法。

(5)
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

解法一

思路

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猜中公平骰子点数的概率为1/6，备择为更大。求至少猜中5次的精确二项右尾概率。

答题过程

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Let pp be the probability that Arvind predicts a score correctly.

H0:p=16,H1:p>16.H_0:p=\frac16,\qquad H_1:p>\frac16.

Under H0H_0, the number CC of correct predictions has distribution B(15,16)\mathrm{B}(15,\frac16). For the observed count of 5,

 P(C≥5∣H0)= 1−(56)15 −15(16)(56)14 −(152)(16)2(56)13 −(153)(16)3(56)12 −(154)(16)4(56)11= 0.0897657….\begin{align*} &\,\mathrm{P}(C\geq5\mid H_0)\\[4mm] =&\,1-\bigg(\frac56\bigg)^{15}\\[4mm] &\,-15\bigg(\frac16\bigg)\bigg(\frac56\bigg)^{14}\\[4mm] &\,\hspace{2pt}-\binom{15}{2}\bigg(\frac16\bigg)^2\bigg(\frac56\bigg)^{13}\\[4mm] &\,\hspace{4pt}-\binom{15}{3}\bigg(\frac16\bigg)^3\bigg(\frac56\bigg)^{12}\\[4mm] &\,\hspace{6pt}-\binom{15}{4}\bigg(\frac16\bigg)^4\bigg(\frac56\bigg)^{11}\\[4mm] =&\,0.0897657\ldots. \end{align*}

Since 0.0897657…<0.100.0897657\ldots<0.10, we reject H0H_0. There is sufficient evidence at the 10% significance level to support Arvind’s claim that he predicts correctly more often than by guessing.