题目

Problem

The heights, in cm, of 200 adults in Barimba are summarised in the following table.

Height ( $h\,\mathrm{cm}$ )	$130\leq h<150$	$150\leq h<160$	$160\leq h<170$	$170\leq h<175$	$175\leq h<195$
Frequency	16	32	76	64	12

(a) Draw a histogram to represent this information.

(4)

(b) The interquartile range is $R\,\mathrm{cm}$ . Show that $R$ is not greater than $15$ .

(2)

题目中文翻译

Barimba 的 200 名成年人的身高（以厘米为单位）汇总如下表。

身高（ $h\,\mathrm{cm}$ ）	$130\leq h<150$	$150\leq h<160$	$160\leq h<170$	$170\leq h<175$	$175\leq h<195$
频数	16	32	76	64	12

(a) 画一幅直方图表示这些数据。

(4)

(b) 四分位距为 $R\,\mathrm{cm}$ 。证明 $R$ 不大于 $15$ 。

(2)

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

组距不等，柱高必须用频数密度，即频数除以组距。

答题过程

展开

The frequency density is the frequency divided by the class width.

Height interval (cm)	Class width	Frequency density
$130\leq h<150$	$20$	$16/20=0.8$
$150\leq h<160$	$10$	$32/10=3.2$
$160\leq h<170$	$10$	$76/10=7.6$
$170\leq h<175$	$5$	$64/5=12.8$
$175\leq h<195$	$20$	$12/20=0.6$

Draw adjoining bars over these intervals, with these densities as their heights. Label the horizontal axis “Height (cm)” and the vertical axis “Frequency density”.

(b)

解法一

思路

展开

先找第50与第150个数据所在组，利用四分位数的上下界证明，不必作均匀分布插值。

答题过程

展开

The cumulative frequencies are $16,48,124,188,200$ .

The lower quartile is in the class $160\leq h<170$ , while the upper quartile is in $170\leq h<175$ . Thus

Q_1\geq160,\qquad Q_3<175.

Consequently,

R=Q_3-Q_1<175-160=15.

Therefore $R$ is not greater than 15 cm.

CIE 9709 2024 June Paper 51 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