题目

Problem

The weights, $x\,\mathrm{kg}$ , of 120 students in a sports college are recorded. The results are summarised in the following table.

Weight ( $x\,\mathrm{kg}$ )	$x\leq 40$	$x\leq 60$	$x\leq 65$	$x\leq 70$	$x\leq 85$	$x\leq 100$
Cumulative frequency	0	14	38	60	106	120

(a) Draw a cumulative frequency graph to represent this information.

(2)

(b) It is found that 35% of the students weigh more than $W\,\mathrm{kg}$ .

Use your graph to estimate the value of $W$ .

(2)

(6)

题目中文翻译

记录某体育学院 120 名学生的体重 $x\,\mathrm{kg}$ ，结果汇总于下表。

体重（ $x\,\mathrm{kg}$ ）	$x\leq 40$	$x\leq 60$	$x\leq 65$	$x\leq 70$	$x\leq 85$	$x\leq 100$
累积频数	0	14	38	60	106	120

(a) 画一条累积频数曲线表示这些数据。

(2)

(b) 发现 35% 的学生体重大于 $W\,\mathrm{kg}$ 。

利用你所画的图估计 $W$ 的值。

(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

(a)

解法一

思路

展开

已经给累计频数，按各重量上界描点，从(40,0)画平滑递增曲线。

答题过程

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Plot the points

(40,0),\ (60,14),\ (65,38),\ (70,60),\ (85,106),\ (100,120).

Use linear scales with weight (kg) on the horizontal axis and cumulative frequency on the vertical axis. Join the points with a smooth increasing curve starting at $(40,0)$ .

(b)

解法一

思路

展开

35%高于W，故W对应65%的累计人数，即78。

答题过程

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The required cumulative frequency is

120(1-0.35)=78.

Read horizontally from cumulative frequency 78 to the curve and then vertically down to the weight axis. This gives $W\approx76\,\mathrm{kg}$ .

(c)

解法一

思路

展开

累计频数作差求实际组频数；40以下人数为0，无须给该空组虚构中点。

答题过程

展开

Subtract successive cumulative frequencies to obtain the class frequencies.

Weight interval (kg)	Frequency $f$	Midpoint $m$
$40<x\leq60$	$14$	$50$
$60<x\leq65$	$24$	$62.5$
$65<x\leq70$	$22$	$67.5$
$70<x\leq85$	$46$	$77.5$
$85<x\leq100$	$14$	$92.5$

There are no students weighing 40 kg or less. Hence

\begin{align*} \sum fm=&\,14(50)+24(62.5)+22(67.5)\\[4mm] &\,\hspace{2pt}+46(77.5)+14(92.5)=8545, \end{align*}

and

\begin{align*} \sum fm^2=&\,14(50)^2+24(62.5)^2+22(67.5)^2\\[4mm] &\,\hspace{2pt}+46(77.5)^2+14(92.5)^2\\[4mm] =&\,625062.5. \end{align*}

The estimated mean is

\overline x=\frac{8545}{120}=71.208333\ldots\approx\boxed{71.2\text{ kg}}.

The estimated standard deviation is

\begin{align*} \sigma=&\,\sqrt{\frac{625062.5}{120}-\biggl(\frac{8545}{120}\biggr)^2}\\[4mm] =&\,\sqrt{138.227430\ldots}\\[4mm] =&\,11.757016\ldots\approx\boxed{11.8\text{ kg}}. \end{align*}

CIE 9709 2023 November Paper 53 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

(c)

解法一

思路

答题过程