题目

Problem

On a certain day, the heights of 150 sunflower plants grown by children at a local school are measured, correct to the nearest cm. These heights are summarised in the following table.

Height (cm)	10–19	20–29	30–39	40–44	45–49	50–54	55–59
Frequency	10	18	32	42	28	14	6

(a) Draw a cumulative frequency graph to illustrate the data.

(4)

(b) Use your graph to estimate the 30th percentile of the heights of the sunflower plants.

(2)

(5)

题目中文翻译

某一天，测量当地一所学校的孩子们种植的 150 株向日葵的高度，精确到最接近的厘米。这些高度汇总如下。

高度（cm）	10–19	20–29	30–39	40–44	45–49	50–54	55–59
频数	10	18	32	42	28	14	6

(a) 画一幅累积频数图表示这些数据。

(4)

(b) 利用所画的图，估计向日葵高度的第 30 百分位数。

(2)

(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

展开

高度记录到最近整数厘米，使用连续组上界19.5等，并累加频数，从最低组界9.5起画平滑曲线。

答题过程

展开

The cumulative frequencies are $10,28,60,102,130,144,150$ .

Plot

(9.5,0),\ (19.5,10),\ (29.5,28),\ (39.5,60),

(44.5,102),\ (49.5,130),\ (54.5,144),\ (59.5,150).

Use linear scales with height (cm) on the horizontal axis and cumulative frequency on the vertical axis. Join the points with a smooth increasing curve, starting at $(9.5,0)$ and ending at $(59.5,150)$ .

(b)

解法一

思路

展开

第30百分位数对应150株向日葵的30%，从累计频数45读高度。

答题过程

展开

The required cumulative frequency is

0.30(150)=45.

Draw a horizontal line from cumulative frequency 45 to the curve, then read down to the height axis. The estimated 30th percentile is 36 cm.

(c)

解法一

思路

展开

使用组中点估计各组数据，分别累计fm与fm²。标准差使用未舍入均值。

答题过程

展开

The class midpoints and weighted values are

$m$	$f$	$fm$	$fm^2$
$14.5$	$10$	$145$	$2102.5$
$24.5$	$18$	$441$	$10804.5$
$34.5$	$32$	$1104$	$38088$
$42$	$42$	$1764$	$74088$
$47$	$28$	$1316$	$61852$
$52$	$14$	$728$	$37856$
$57$	$6$	$342$	$19494$
Total	$150$	$5840$	$244285$

Therefore the estimated mean is

\overline h=\frac{5840}{150}=38.933333\ldots \approx\boxed{38.9\text{ cm}}.

The estimated standard deviation is

\begin{align*} \sigma=&\,\sqrt{\frac{244285}{150}-\biggl(\frac{5840}{150}\biggr)^2}\\[4mm] =&\,\sqrt{112.762222\ldots}\\[4mm] =&\,10.618955\ldots\\[4mm] \approx&\,\boxed{10.6\text{ cm}}. \end{align*}

CIE 9709 2024 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