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IAL 2024 Jan Q1

A Level / Edexcel / S1

IAL 2024 Jan Paper · Question 1

题目

Problem

The histogram below shows the distribution of the heights, to the nearest cm, of 408408 plants.

Histogram showing heights of plants

(a) Use the histogram to complete the following table.

Height (h cm)(h\text{ cm})5⩽h<95\leqslant h<99⩽h<139\leqslant h<1313⩽h<1513\leqslant h<1515⩽h<1715\leqslant h<1717⩽h<2517\leqslant h<25
Frequency3232152152120120
(2)

(b) Use interpolation to estimate the median.

(2)

The mean height of these plants is 13.213.2 cm correct to one decimal place.

(c) Describe the skew of these data. Give a reason for your answer.

(1)

Two of these plants are chosen at random.

(d) Estimate the probability that both of their heights are between 88 cm and 1414 cm.

(3)

解答

(a)

解法一

思路

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Histogram 的面积代表频数。从图上对应两组面积可得到剩下两个频数。已知总频数是 408408，也可检查两格合计应为 104104。

答题过程

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From the histogram, the missing frequencies are

72and32.\begin{align*} 72\quad\text{and}\quad32. \end{align*}

So the completed table is

Height (h cm)(h\text{ cm})5⩽h<95\leqslant h<99⩽h<139\leqslant h<1313⩽h<1513\leqslant h<1515⩽h<1715\leqslant h<1717⩽h<2517\leqslant h<25
Frequency323215215212012072723232

(b)

解法一

思路

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共有 408408 个数据，中位数位置是第 204204 个。到 1313 cm 前累计频数是 184184，到 1515 cm 前累计频数是 304304，所以中位数在 13⩽h<1513\leqslant h<15 这一组。

答题过程

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The median is the 204204th value.

The cumulative frequency before 1313 cm is

32+152=184.\begin{align*} 32+152=184. \end{align*}

The median lies in the class 13⩽h<1513\leqslant h<15.

Using interpolation,

median= 13+204−184120(15−13)= 13+20120(2)= 13.333….\begin{align*} \text{median} =&\,13+\frac{204-184}{120}(15-13)\\[3mm] =&\,13+\frac{20}{120}(2)\\[3mm] =&\,13.333\ldots. \end{align*}

So the estimated median is

13.3 cm.\begin{align*} 13.3\text{ cm}. \end{align*}

(c)

解法一

思路

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平均数约为 13.213.2，中位数约为 13.313.3，两者很接近，所以可认为近似对称、没有明显 skew。

答题过程

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The data are approximately symmetric, since the mean 13.213.2 cm is close to the estimated median 13.313.3 cm.

(d)

解法一

思路

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先估计 88 到 1414 cm 的植物数量。它包含 88 到 99 的一部分、99 到 1313 整组、1313 到 1414 的一部分。抽两株不放回，所以第二个分母是 407407。

答题过程

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Estimated number between 88 cm and 1414 cm:

n= 14(32)+152+12(120)= 8+152+60= 220.\begin{align*} n =&\,\frac14(32)+152+\frac12(120)\\[3mm] =&\,8+152+60\\[3mm] =&\,220. \end{align*}

Therefore the required probability is

220408×219407= 0.2901….\begin{align*} \frac{220}{408}\times\frac{219}{407} =&\,0.2901\ldots. \end{align*}

So the probability is approximately

0.290.\begin{align*} 0.290. \end{align*}