CalcGospel 國際數學圖譜
返回

S1 2021 June Q3

A Level / Edexcel / S1

IAL 2021 June Paper · Question 3

题目

Problem

A random sample of 100 carrots is taken from a farm and their lengths, LL cm, recorded. The data are summarised in the following table.

Length, LL cmFrequency, ffClass mid point, xx cm
5≤L<85\leq L<856.5
8≤L<108\leq L<10139
10≤L<1210\leq L<121611
12≤L<1512\leq L<152513.5
15≤L<2015\leq L<203017.5
20≤L<2820\leq L<281124

A histogram is drawn to represent these data.

The bar representing the class 5≤L<85\leq L<8 is 1.5 cm wide and 1 cm high.

(a) Find the width and height of the bar representing the class 15≤L<2015\leq L<20

(3)

(b) Use linear interpolation to estimate the median length of these carrots.

(2)

(c) Estimate

(i) the mean length of these carrots,

(2)

(ii) the standard deviation of the lengths of these carrots.

(3)

A supermarket will only buy carrots with length between 9 cm and 22 cm.

(d) Estimate the proportion of carrots from the farm that the supermarket will buy.

(2)

Any carrots that the supermarket does not buy are sold as animal feed. The farm makes a profit of 2.2 pence on each carrot sold to the supermarket, a profit of 0.8 pence on each carrot longer than 22 cm and a loss of 1.2 pence on each carrot shorter than 9 cm.

(e) Find an estimate of the mean profit per carrot made by the farm.

(2)
题目中文翻译

从一个农场随机抽取 100 根胡萝卜，并记录它们的长度 LL cm。数据汇总如下表。

长度，LL cm频数，ff组中值，xx cm
5≤L<85\leq L<856.5
8≤L<108\leq L<10139
10≤L<1210\leq L<121611
12≤L<1512\leq L<152513.5
15≤L<2015\leq L<203017.5
20≤L<2820\leq L<281124

用直方图表示这些数据。表示 5≤L<85\leq L<8 这一组的柱宽为 1.5 cm、高为 1 cm。

(a) 求表示 15≤L<2015\leq L<20 这一组的柱宽和柱高。

(b) 使用线性插值估计这些胡萝卜长度的中位数。

(c) 估计：

(i) 这些胡萝卜的平均长度；

(ii) 这些胡萝卜长度的标准差。

一家超市只购买长度在 9 cm 到 22 cm 之间的胡萝卜。

(d) 估计农场中会被该超市购买的胡萝卜所占比例。

超市不购买的胡萝卜会作为动物饲料出售。每根卖给超市的胡萝卜为农场带来 2.2 便士利润；每根长于 22 cm 的胡萝卜带来 0.8 便士利润；每根短于 9 cm 的胡萝卜造成 1.2 便士亏损。

(e) 估计农场每根胡萝卜的平均利润。

解答

(a)

解法一

思路

展开

直方图中面积代表频数。已知第一组的频数 5 对应图上面积 1.5×1=1.51.5\times1=1.5，所以频数 30 对应的面积是 30÷530\div5 倍。再用新组距的图上宽度求高度。

答题过程

展开

The class width 5≤L<85\leq L<8 is 3 cm and is drawn as 1.5 cm, so the scale is

3 cm in data=1.5 cm on the histogram.\begin{align*} 3\text{ cm in data}=1.5\text{ cm on the histogram}. \end{align*}

For 15≤L<2015\leq L<20, the class width is 5 cm, so the width of the bar is

5×1.53=2.5 cm.\begin{align*} 5\times\frac{1.5}{3}=2.5\text{ cm}. \end{align*}

The first bar has area

1.5×1=1.5,\begin{align*} 1.5\times1=1.5, \end{align*}

and this represents frequency 5. The frequency 30 bar must therefore have area

1.5×305=9.\begin{align*} 1.5\times\frac{30}{5}=9. \end{align*}

Hence its height is

92.5=3.6 cm.\begin{align*} \frac{9}{2.5}=3.6\text{ cm}. \end{align*}

(b)

解法一

思路

展开

总频数是 100，所以中位数在第 50 个数据附近。累计频数到 10≤L<1210\leq L<12 为 34，因此中位数落在 12≤L<1512\leq L<15 这一组。

答题过程

展开

The cumulative frequency before the class 12≤L<1512\leq L<15 is

5+13+16=34.\begin{align*} 5+13+16=34. \end{align*}

The median is the 50th value, so it is 50−34=1650-34=16 values into the class 12≤L<1512\leq L<15.

Using linear interpolation,

median=12+1625×3=13.92.\text{median} =12+\frac{16}{25}\times3 =13.92.

Therefore the median length is approximately

13.9 cm.\begin{align*} 13.9\text{ cm}. \end{align*}

(c)(i)

解法一

思路

展开

分组数据估计平均数时，用组中值代表该组所有数据。

答题过程

展开

Using the class midpoints,

∑fx= 5(6.5)+13(9)+16(11)+25(13.5)+30(17.5)+11(24)= 1452.\begin{align*} \sum fx =&\,5(6.5)+13(9)+16(11)\\ &\quad+25(13.5)+30(17.5)+11(24)\\ =&\,1452. \end{align*}

Hence

xˉ=1452100=14.52.\begin{align*} \bar x=\frac{1452}{100}=14.52. \end{align*}

So the estimated mean length is

14.5 cm.\begin{align*} 14.5\text{ cm}. \end{align*}

(c)(ii)

解法一

思路

展开

继续使用组中值，先求 ∑fx2\sum fx^2，再代入总体标准差公式。

答题过程

展开

Using the class midpoints,

∑fx2= 5(6.52)+13(92)+16(112)+25(13.52)+30(17.52)+11(242)= 23280.\begin{align*} \sum fx^2 =&\,5(6.5^2)+13(9^2)+16(11^2)\\ &\quad+25(13.5^2)+30(17.5^2)+11(24^2)\\ =&\,23280. \end{align*}

Therefore

σ= ∑fx2100−xˉ2= 23280100−14.522= 21.9696= 4.687….\begin{align*} \sigma =&\,\sqrt{\frac{\sum fx^2}{100}-\bar x^2}\\ =&\,\sqrt{\frac{23280}{100}-14.52^2}\\ =&\,\sqrt{21.9696}\\ =&\,4.687\ldots. \end{align*}

So the estimated standard deviation is

4.69 cm.\begin{align*} 4.69\text{ cm}. \end{align*}

(d)

解法一

思路

展开

需要估计 9≤L≤229\leq L\leq22 的比例。把完全包含的组加上，再用线性假设处理边界组：8≤L<108\leq L<10 中取一半，20≤L<2820\leq L<28 中取 2020 到 2222 的四分之一。

答题过程

展开

The estimated number of carrots with 9≤L<229\leq L<22 is

12(13)+16+25+30+28(11).\begin{align*} \frac{1}{2}(13)+16+25+30+\frac{2}{8}(11). \end{align*}

So

6.5+16+25+30+2.75=80.25.\begin{align*} 6.5+16+25+30+2.75=80.25. \end{align*}

Therefore the estimated proportion is

80.25100=0.8025.\begin{align*} \frac{80.25}{100}=0.8025. \end{align*}

(e)

解法一

思路

展开

按 (d) 的估计，买走的是 80.2580.25 个。低于 9 cm 的估计数量是 5+12(13)=11.55+\frac12(13)=11.5，高于 22 cm 的估计数量是 68(11)=8.25\frac68(11)=8.25。分别乘以对应利润，再除以 100。

答题过程

展开

The estimated number of carrots sold to the supermarket is 80.2580.25.

The estimated number shorter than 9 cm is

5+12(13)=11.5.\begin{align*} 5+\frac{1}{2}(13)=11.5. \end{align*}

The estimated number longer than 22 cm is

68(11)=8.25.\begin{align*} \frac{6}{8}(11)=8.25. \end{align*}

So the estimated mean profit per carrot is

2.2(80.25)+0.8(8.25)−1.2(11.5)100= 176.55+6.6−13.8100= 1.6935.\begin{align*} \frac{2.2(80.25)+0.8(8.25)-1.2(11.5)}{100} =&\,\frac{176.55+6.6-13.8}{100}\\ =&\,1.6935. \end{align*}

Therefore the estimated mean profit is

1.7 pence per carrot.\begin{align*} 1.7\text{ pence per carrot}. \end{align*}