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CIE 9709 2023 June Paper 61 Q6

A Level / CIE / S2

CIE 9709 2023 June Paper 61 Paper · Question 6

题目

Problem

A sample of 55 randomly selected values of a variable XX is as follows:

1261a1\qquad2\qquad6\qquad1\qquad a

where a>0a>0.

Given that an unbiased estimate of the variance of XX calculated from this sample is 112\dfrac{11}{2}, find the value of aa.

(3)
题目中文翻译

从变量 XX 中随机抽取 55 个取值作为样本,结果如下。

1261a1\qquad2\qquad6\qquad1\qquad a

其中 a>0a>0。

已知根据该样本计算出的 XX 的方差的一个无偏估计值为 112\dfrac{11}{2},求 aa 的值。

(3)
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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以a表达总和与平方和,代入无偏方差公式,解出两根后用a大于0排除零。

答题过程

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For the sample,

∑x=10+a,∑x2=42+a2.\sum x=10+a,\qquad\sum x^2=42+a^2.

Using the unbiased variance,

14(42+a2−(10+a)25)=112.\frac14\bigg(42+a^2-\frac{(10+a)^2}{5}\bigg)=\frac{11}{2}.

Thus

210+5a2−(100+20a+a2)=110,210+5a^2-(100+20a+a^2)=110,

giving

4a2−20a=0⟹a(a−5)=0.4a^2-20a=0 \quad\Longrightarrow\quad a(a-5)=0.

Since a>0a>0, a=5\boxed{a=5}.