CalcGospel 國際數學圖譜
返回

CIE 9709 2024 June Paper 62 Q2

A Level / CIE / S2

CIE 9709 2024 June Paper 62 Paper · Question 2

题目

Problem

Henri wants to choose a random sample from the 804804 students at his college. He numbers the students from 11 to 804804 and then uses random numbers generated by his calculator. The first 2020 random digits produced by his calculator are as follows.

5  6  7  1  0  9  8  4  3  1  0  9  6  6  5  0  2  1  7  65\;6\;7\;1\;0\;9\;8\;4\;3\;1\;0\;9\;6\;6\;5\;0\;2\;1\;7\;6

Henri’s first two student numbers are 567567 and 109109.

(a) Use Henri’s digits to find the numbers of the next two students in the sample.

(2)

There were 3030 students in Henri’s sample. He asked each of them how much time, XX hours, they spent on social media each week, on average. He summarised the results as follows.

n=30∑x=610∑x2=12 405n=30\qquad\sum x=610\qquad\sum x^2=12\,405

(b) Use this information to calculate an unbiased estimate of the mean of XX and show that an unbiased estimate of the variance of XX is less than 0.10.1.

(3)

(c) Henri’s friend claims that Henri has probably made a mistake in his calculation of ∑x\sum x or ∑x2\sum x^2.

Use your answer to part (b) to comment on this claim.

(1)
题目中文翻译

Henri 想从他所在学院的 804804 名学生中抽取一个随机样本。他给学生编上从 11 到 804804 的号码，然后使用计算器生成的随机数。计算器最先生成的 2020 个随机数字如下。

5  6  7  1  0  9  8  4  3  1  0  9  6  6  5  0  2  1  7  65\;6\;7\;1\;0\;9\;8\;4\;3\;1\;0\;9\;6\;6\;5\;0\;2\;1\;7\;6

Henri 最先选出的两名学生的编号是 567567 和 109109。

(a) 用 Henri 的这些数字求样本中接下来两名学生的编号。

(2)

Henri 的样本中有 3030 名学生。他询问每人平均每周在社交媒体上花多少时间，记为 XX 小时。他将结果汇总如下。

n=30∑x=610∑x2=12 405n=30\qquad\sum x=610\qquad\sum x^2=12\,405

(b) 利用这些信息，计算 XX 的均值的一个无偏估计值，并证明 XX 的方差的一个无偏估计值小于 0.10.1。

(3)

(c) Henri 的朋友认为，Henri 可能算错了 ∑x\sum x 或 ∑x2\sum x^2。

根据 (b) 的结果评价这一说法。

(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

每三个数字为一组，拒绝超出1至804范围的编号。不要重叠取组。

答题过程

展开

Grouping the digits in threes gives

567, 109, 843, 109, 665, 021,….567,\ 109,\ 843,\ 109,\ 665,\ 021,\ldots.

Reject 843 because it exceeds 804, and reject the repeated 109. The next two student numbers are 665 and 21\boxed{665\text{ and }21}.

(b)

解法一

思路

展开

用样本均值和无偏方差公式，保留精确分数以证明确实小于0.1。

答题过程

展开

The unbiased estimate of the population mean is

xˉ= 61030=613≈ 20.3 hours(3 s.f.).\begin{align*} \bar x=&\,\frac{610}{30}=\frac{61}{3}\\[4mm] \approx&\,\boxed{20.3\text{ hours}}\quad\text{(3 s.f.).} \end{align*}

The unbiased estimate of the population variance is

s2= 12405−6102/3029= 587= 0.0574713…<0.1,\begin{align*} s^2=&\,\frac{12405-610^2/30}{29}\\[4mm] =&\,\frac{5}{87}\\[4mm] =&\,0.0574713\ldots<0.1, \end{align*}

as required.

(c)

解法一

思路

展开

方差约0.0575表示30人的每周使用时间极接近，结合社交媒体用时情境评论其合理性。

答题过程

展开

The claim is plausible. The estimated variance is unrealistically small for weekly social-media times, corresponding to a standard deviation of only about 0.240 hours, so Henri has probably made a mistake.