题目

Problem

The times, $T$ minutes, taken by a random sample of $75$ students to complete a test were noted. The results were summarised by $\sum t=230$ and $\sum t^2=930$ .

(a) Calculate unbiased estimates of the population mean and variance of $T$ .

(3)

You should now assume that your estimates from part (a) are the true values of the population mean and variance of $T$ .

(b) The times taken by another random sample of $75$ students were noted, and the sample mean, $\overline{T}$ , was found.

Find the value of $a$ such that $P(\overline{T}>a)=0.234$ .

(3)

题目中文翻译

随机抽取 $75$ 名学生，记录他们完成一次测试所用的时间，记为 $T$ 分钟。汇总结果为 $\sum t=230$ 和 $\sum t^2=930$ 。

(a) 计算 $T$ 的总体均值和方差的无偏估计值。

(3)

现在假定 (a) 中求得的估计值就是 $T$ 的总体均值和方差的真值。

(b) 另随机抽取 $75$ 名学生，记录他们的用时，并求得样本均值 $\overline{T}$ 。

求 $a$ 的值，使 $P(\overline{T}>a)=0.234$ 。

(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

(a)

解法一

思路

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样本均值是总体均值的无偏估计，样本离均差平方和除以74给出总体方差的无偏估计。

答题过程

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The unbiased estimate of the population mean is

\begin{align*} \bar t=&\,\frac{230}{75}=\frac{46}{15}\\[4mm] \approx&\,\boxed{3.07\text{ minutes}}\quad\text{(3 s.f.).} \end{align*}

The unbiased estimate of the population variance is

\begin{align*} s^2=&\,\frac{930-230^2/75}{74}\\[4mm] =&\,\frac{337}{111}\\[4mm] =&\,\boxed{3.04\text{ minutes}^2}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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按题意将估计值当作总体真值，中心极限定理给出样本均值的近似分布；上尾0.234对应累计概率0.766。

答题过程

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Using the unrounded values from part (a), the Central Limit Theorem gives

\overline T\approx\mathrm{N}\bigg(\frac{46}{15},\frac{337}{111(75)}\bigg).

As $\mathrm{P}(\overline T>a)=0.234$ ,

\begin{align*} \frac{a-46/15}{\sqrt{337/(111(75))}}=&\,\Phi^{-1}(0.766)\\[4mm] =&\,0.7257\ldots. \end{align*}

Therefore

\begin{align*} a=&\,\frac{46}{15}+\Phi^{-1}(0.766)\sqrt{\frac{337}{111(75)}}\\[4mm] =&\,3.21268\ldots, \end{align*}

giving $\boxed{a=3.21\text{ minutes}}$ (3 s.f.).

CIE 9709 2024 November Paper 61 Q3

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程