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CIE 9231 2024 June Paper 43 Q2

A Level / CIE / FS

CIE 9231 2024 June Paper 43 Paper · Question 2

题目

Problem

A rowing club has a large number of members. A random sample of 12 of these members is taken and the pulse rate, xx beats per minute (bpm), of each is measured after a 30-minute training session. A 98% confidence interval for the population mean pulse rate, μ\mu bpm, is calculated from the sample as 64.22<μ<68.6664.22 < \mu < 68.66.

(a) Find the values of ∑x\sum x and ∑x2\sum x^2.

[6]

(b) State an assumption that is necessary for the confidence interval to be valid.

[1]
题目中文翻译

一个赛艇俱乐部有大量会员。从这些会员中随机抽取 12 人,并测量每人在一次 30 分钟训练后的脉搏率 xx,单位为每分钟跳动次数(bpm)。根据该样本计算出总体平均脉搏率 μ\mu bpm 的 98% 置信区间为 64.22<μ<68.6664.22 < \mu < 68.66。

(a) 求 ∑x\sum x 和 ∑x2\sum x^2 的值。

(6)

(b) 说明使该置信区间有效所需要的一个假设。

(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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区间中点给出样本均值,半宽给出标准误。用 98%98\% 区间对应的 tt 临界值反求平方和。

答题过程

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The midpoint is (64.22+68.66)/2=66.44(64.22+68.66)/2=66.44, so ∑x=12(66.44)=797.28\sum x=12(66.44)=797.28. The half-width is 2.222.22. With 1111 degrees of freedom, the two-sided 98%98\% critical value is 2.7182.718. Thus

s2=12(2.222.718)2=8.005496….s^2=12\Big(\frac{2.22}{2.718}\Big)^2=8.005496\ldots.

Therefore

∑x2=11s2+797.28212=53059.34…,\sum x^2=11s^2+\frac{797.28^2}{12}=53059.34\ldots,

which is approximately 5310053100 to three significant figures.

(b)

解法一

思路

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小样本 tt 区间所需的正态性针对训练后的总体脉搏率。

答题过程

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The population of pulse rates after the training session is assumed to be normally distributed.