题目

Problem

Shane is studying the lengths of the tails of male red kangaroos. He takes a random sample of 14 male red kangaroos and measures the length of the tail, $x$ m, for each kangaroo. He then calculates a 90% confidence interval for the population mean tail length, $\mu$ m, of male red kangaroos. He assumes that the tail lengths are normally distributed and finds that $1.11 \leq \mu \leq 1.14$ .

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

[6]

题目中文翻译

Shane 正在研究雄性红袋鼠尾巴的长度。他随机抽取 14 只雄性红袋鼠，并测量每只袋鼠尾巴的长度 $x$ m。然后，他计算出雄性红袋鼠总体平均尾长 $\mu$ m 的 90% 置信区间。他假设尾巴长度服从正态分布，并得到 $1.11 \leq \mu \leq 1.14$ 。

求该样本的 $\sum x$ 和 $\sum x^2$ 的值。

(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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用区间的中点反推样本均值，再用半宽和临界值求无偏方差。

答题过程

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The sample mean is $(1.11+1.14)/2=1.125$ , so $\sum x=14(1.125)=15.75$ . The half-width is $0.015$ . The two-sided $90\%$ critical value with $13$ degrees of freedom is $1.771$ , giving

s^2=14\Big(\frac{0.015}{1.771}\Big)^2=0.00100432\ldots.

Therefore

\sum x^2=13s^2+\frac{15.75^2}{14}=17.73180\ldots,

or $17.7$ to three significant figures.

CIE 9231 2023 June Paper 43 Q2

题目

解答

解法一

思路

答题过程