题目

Problem

In a certain country, the wingspan, in cm, of adult swans has the distribution $\mathrm{N}(234.0,6.9^2)$ .

(a) Find the probability that the mean wingspan for a random sample of $25$ adult swans is between $232.0\,\mathrm{cm}$ and $235.0\,\mathrm{cm}$ .

(4)

A naturalist wishes to test whether the mean wingspan of adult swans in a particular region is different from $234.0\,\mathrm{cm}$ . He chooses a random sample of $25$ swans in this region and finds that the mean wingspan is $236.3\,\mathrm{cm}$ . Assume that the standard deviation of the wingspans is $6.9\,\mathrm{cm}$ .

(b) (i) Carry out the test at the $5\%$ significance level.

(5)

(ii) Explain whether it was necessary to use the Central Limit Theorem in 6(b)(i).

(1)

题目中文翻译

在某个国家，成年天鹅的翼展（以厘米为单位）服从分布 $\mathrm{N}(234.0,6.9^2)$ 。

(a) 求从成年天鹅中随机抽取 $25$ 只，所得平均翼展介于 $232.0\,\mathrm{cm}$ 和 $235.0\,\mathrm{cm}$ 之间的概率。

(4)

一名博物学家想检验某地区成年天鹅的平均翼展是否与 $234.0\,\mathrm{cm}$ 不同。他从该地区随机抽取 $25$ 只天鹅，发现平均翼展为 $236.3\,\mathrm{cm}$ 。假设翼展的标准差为 $6.9\,\mathrm{cm}$ 。

(b) (i) 在 $5\%$ 的显著性水平下进行检验。

(5)

(ii) 说明在 6(b)(i) 中是否有必要使用中心极限定理。

(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

样本均值的方差为总体方差除以样本量，标准差是 $6.9/5$ ；总体已经正态，所以无需近似或连续性修正。

答题过程

展开

Let $\overline W$ be the sample mean wingspan. Since the population is normal,

\begin{align*} \overline W\sim&\,\mathrm N\bigg(234,\frac{6.9^2}{25}\bigg),\\[4mm] \operatorname{sd}(\overline W)=&\,\frac{6.9}{5}=1.38. \end{align*}

Therefore

\begin{align*} &\,\mathrm P(232<\overline W<235)\\[4mm] =&\,\Phi\bigg(\frac{235-234}{1.38}\bigg)\\[4mm] &\,-\Phi\bigg(\frac{232-234}{1.38}\bigg)\\[4mm] =&\,\Phi(0.724638\ldots)-\Phi(-1.449275\ldots)\\[4mm] \approx&\,\boxed{0.692}. \end{align*}

(b)(i)

解法一

思路

展开

怀疑“不同”对应双侧检验；结论只说明证据不足，不能把未拒绝原假设写成证明均值相等。

答题过程

展开

Let $\mu$ be the population mean wingspan of adult swans in this region. Test

H_0:\mu=234,\qquad H_1:\mu\ne234.

Using the known standard deviation $6.9$ and sample size $25$ , the observed statistic is

z=\frac{236.3-234}{6.9/\sqrt{25}}=1.66667\ldots.

The critical values for a two-tailed $5\%$ test are $\pm1.96$ . Since $|1.66667|<1.96$ , do not reject $H_0$ .

There is insufficient evidence at the $5\%$ significance level that the population mean wingspan of adult swans in this region differs from $234\ \text{cm}$ .

(b)(ii)

解法一

思路

展开

总体正态直接保证样本均值正态，不以样本量是否大于某个经验门槛为理由。

答题过程

展开

No. The population distribution of adult-swan wingspans is normal, so the sample mean is normal without needing the Central Limit Theorem.

CIE 9709 2026 June Paper 65 Q6

题目

解答

(a)

解法一

思路

答题过程

(b)(i)

解法一

思路

答题过程

(b)(ii)

解法一

思路

答题过程