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CIE 9709 2022 March Paper 62 Q1

A Level / CIE / S2

CIE 9709 2022 March Paper 62 Paper · Question 1

题目

Problem

The lengths, in millimetres, of a random sample of 1212 rods made by a certain machine are as follows.

200  201  198  202  200  199  199  201  197  202  200  199200\;201\;198\;202\;200\;199\;199\;201\;197\;202\;200\;199

(a) Find unbiased estimates of the population mean and variance.

(3)

(b) Give a statistical reason why these estimates may not be reliable.

(1)
题目中文翻译

从某台机器制造的杆件中随机抽取 1212 根,其长度(以毫米为单位)如下。

200  201  198  202  200  199  199  201  197  202  200  199200\;201\;198\;202\;200\;199\;199\;201\;197\;202\;200\;199

(a) 求总体均值和方差的无偏估计值。

(3)

(b) 从统计学角度给出一个理由,说明为什么这些估计值可能不可靠。

(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

先求总和与平方和,均值直接除以12;无偏方差分母用11。

答题过程

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The sample has n=12n=12, ∑x=2398\sum x=2398 and ∑x2=479226\sum x^2=479226. The unbiased estimate of the population mean is

x‾=239812=11996 mm.\overline x=\frac{2398}{12}=\boxed{\frac{1199}{6}\text{ mm}}.

The unbiased estimate of the population variance is

s2= 111(479226−2398212)= 73 mm2.\begin{align*} s^2=&\,\frac{1}{11}\bigg(479226-\frac{2398^2}{12}\bigg)\\[4mm] =&\,\boxed{\frac73\text{ mm}^2}. \end{align*}

(b)

解法一

思路

展开

样本仅12根,样本量较小使估计不够可靠。

答题过程

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The sample size is small, with only 12 rods, so the estimates may not be reliable.