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CIE 9231 2022 Nov Paper 42 Q1

A Level / CIE / FS

CIE 9231 2022 Nov Paper 42 Paper · Question 1

题目

Problem

A basketball club has a large number of players. The heights, xx m, of a random sample of 10 of these players are measured. A 90% confidence interval for the population mean height, μ\mu m, of players in this club is calculated. It is assumed that heights are normally distributed. The confidence interval is 1.78≤μ≤2.021.78 \leq \mu \leq 2.02.

Find the values of ∑x\sum x and ∑x2\sum x^2 for this sample.

[6]
题目中文翻译

一个篮球俱乐部有大量球员。随机抽取其中 10 名球员,测量他们的身高,记为 xx m。假设身高服从正态分布,并计算该俱乐部球员总体平均身高 μ\mu m 的 90% 置信区间。该置信区间为 1.78≤μ≤2.021.78 \leq \mu \leq 2.02。

求这组样本的 ∑x\sum x 和 ∑x2\sum x^2。

(6)
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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置信区间的中点是样本均值,半宽是临界值乘标准误。先倒推出样本方差,再由无偏方差公式求平方和。

答题过程

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The midpoint is xˉ=(1.78+2.02)/2=1.90\bar x=(1.78+2.02)/2=1.90, so ∑x=10(1.90)=19.0\sum x=10(1.90)=19.0. The half-width is 0.120.12. Using the 90%90\% critical value t0.95,9=1.833t_{0.95,9}=1.833 gives

s2=10(0.121.833)2.s^2=10\Big(\frac{0.12}{1.833}\Big)^2.

Since 9s2=∑x2−(∑x)2/109s^2=\sum x^2-(\sum x)^2/10, we get

∑x2=9s2+19210=36.48572…≈36.5.\sum x^2=9s^2+\frac{19^2}{10}=36.48572\ldots\approx36.5.