题目

Problem

The times taken, $t$ minutes, by 300 students to travel to Hollowton College are recorded. The results are summarised in the table below.

Time ( $t$ minutes)	$t\leq 10$	$t\leq 20$	$t\leq 30$	$t\leq 40$	$t\leq 60$	$t\leq 90$
Cumulative frequency	34	86	142	208	265	300

(a) On the grid, draw a cumulative frequency graph to illustrate this information.

(2)

(b) 120 students take more than $k$ minutes to travel to college. Use your graph to estimate the value of $k$ .

(2)

(6)

题目中文翻译

记录了 300 名学生前往 Hollowton College 所用的时间 $t$ 分钟。结果汇总如下表。

时间（ $t$ 分钟）	$t\leq 10$	$t\leq 20$	$t\leq 30$	$t\leq 40$	$t\leq 60$	$t\leq 90$
累积频数	34	86	142	208	265	300

(a) 在网格上画出累积频数图来表示这些数据。

(2)

(b) 120 名学生前往学校所用的时间超过 $k$ 分钟。利用你所画的图估计 $k$ 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

表中是累计频数，直接在时间上界描点，从原点画平滑递增曲线。

答题过程

展开

Plot the points

(0,0),\ (10,34),\ (20,86),\ (30,142),

(40,208),\ (60,265),\ (90,300).

Join them with a smooth increasing curve. Label the axes “Time, $t$ (minutes)” and “Cumulative frequency”.

(b)

解法一

思路

展开

120人超过k分钟，表示180人在k分钟或以下。因此从累计频数180处读图。

答题过程

展开

The cumulative frequency at $k$ is $300-120=180$ .

Draw a horizontal line from cumulative frequency 180 to the curve, then read down to the time axis. This gives an estimate of $k=35$ minutes.

(c)

解法一

思路

展开

累计频数作差才是各组频数，再以组中点计算均值与二阶矩，最后求标准差。

答题过程

展开

The frequencies are the successive differences of the cumulative frequencies:

34,\ 86-34,\ 142-86,\ 208-142,\ 265-208,\ 300-265.

Thus

Interval (minutes)	Frequency, $f$	Midpoint, $m$
$0<t\leq10$	$34$	$5$
$10<t\leq20$	$52$	$15$
$20<t\leq30$	$56$	$25$
$30<t\leq40$	$66$	$35$
$40<t\leq60$	$57$	$50$
$60<t\leq90$	$35$	$75$

Then

\begin{align*} \sum fm=&\,34(5)+52(15)+56(25)\\[4mm] &\,\hspace{2pt}+66(35)+57(50)+35(75)\\[4mm] =&\,10135, \end{align*}

and

\begin{align*} \sum fm^2=&\,34(5^2)+52(15^2)+56(25^2)\\[4mm] &\,\hspace{2pt}+66(35^2)+57(50^2)+35(75^2)\\[4mm] =&\,467775. \end{align*}

The estimated mean is

\begin{align*} \bar t=&\,\frac{10135}{300}=33.78333\ldots\\[4mm] \approx&\,\boxed{33.8\text{ minutes}}. \end{align*}

The estimated standard deviation is

\begin{align*} s=&\,\sqrt{\frac{467775}{300} -\biggl(\frac{10135}{300}\biggr)^2}\\[4mm] =&\,\sqrt{417.936388\ldots}\\[4mm] =&\,20.44349\ldots\\[4mm] \approx&\,\boxed{20.4\text{ minutes}}. \end{align*}

CIE 9709 2025 June Paper 52 Q5

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