题目

Problem

The times, to the nearest minute, of 150 athletes taking part in a charity run are recorded. The results are summarised in the table.

Time in minutes	101–120	121–130	131–135	136–145	146–160
Frequency	18	48	34	32	18

(a) Draw a histogram to represent this information.

(4)

(b) Calculate estimates for the mean and standard deviation of the times taken by the athletes.

(5)

题目中文翻译

记录了 150 名参加慈善跑的运动员所用的时间，精确到最接近的分钟。结果汇总如下表。

时间（分钟）	101–120	121–130	131–135	136–145	146–160
频数	18	48	34	32	18

(a) 画一幅直方图表示这些数据。

(4)

(b) 计算运动员所用时间的均值和标准差的估计值。

(5)

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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时间取整到分钟，连续组界用100.5等。组距不等，柱高必须为频数密度。

答题过程

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The continuous class boundaries and frequency densities are

Time interval (minutes)	Class width	Frequency density
$100.5\leq t<120.5$	$20$	$18/20=0.9$
$120.5\leq t<130.5$	$10$	$48/10=4.8$
$130.5\leq t<135.5$	$5$	$34/5=6.8$
$135.5\leq t<145.5$	$10$	$32/10=3.2$
$145.5\leq t<160.5$	$15$	$18/15=1.2$

Draw adjoining bars over these intervals, with the stated densities as heights. Label the axes “Time (minutes)” and “Frequency density”.

(b)

解法一

思路

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使用连续区间中点，计算加权总和与平方和，方差用未舍入均值。

答题过程

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The class midpoints are $110.5,125.5,133,140.5,153$ minutes. Thus

\begin{align*} \sum fm=&\,18(110.5)+48(125.5)+34(133)\\[4mm] &\,\hspace{2pt}+32(140.5)+18(153)=19785, \end{align*}

and

\begin{align*} \sum fm^2=&\,18(110.5)^2+48(125.5)^2\\[4mm] &\,\hspace{2pt}+34(133)^2+32(140.5)^2\\[4mm] &\,\hspace{4pt}+18(153)^2\\[4mm] =&\,2630272.5. \end{align*}

The estimated mean is

\overline t=\frac{19785}{150}=\boxed{131.9\text{ minutes}},

and the estimated standard deviation is

\begin{align*} \sigma=&\,\sqrt{\frac{2630272.5}{150}-(131.9)^2}\\[4mm] =&\,\sqrt{137.54}=11.727744\ldots\\[4mm] \approx&\,\boxed{11.7\text{ minutes}}. \end{align*}

CIE 9709 2023 November Paper 51 Q4

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程