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S1 2021 Jan Q6

A Level / Edexcel / S1

IAL 2021 Jan Paper · Question 6

题目

Problem

A disc of radius 1 cm is rolled onto a horizontal grid of rectangles so that the disc is equally likely to land anywhere on the grid. Each rectangle is 5 cm long and 3 cm wide. There are no gaps between the rectangles and the grid is sufficiently large so that no discs roll off the grid.

If the disc lands inside a rectangle without covering any part of the edges of the rectangle then a prize is won.

By considering the possible positions for the centre of the disc,

(a) show that the probability of winning a prize on any particular roll is

15\frac15
(3)

A group of 15 students each roll the disc onto the grid twenty times and record the number of times, xx, that each student wins a prize. Their results are summarised as follows

∑x=61∑x2=295\sum x=61\qquad \sum x^2=295

(b) Find the standard deviation of the number of prizes won per student.

(2)

A second group of 12 students each roll the disc onto the grid twenty times and the mean number of prizes won per student is 3.5 with a standard deviation of 2

(c) Find the mean and standard deviation of the number of prizes won per student for the whole group of 27 students.

(7)

The 27 students also recorded the number of times that the disc covered a corner of a rectangle and estimated the probability to be 0.2216 (to 4 decimal places).

(d) Explain how this probability could be used to find an estimate for the value of π\pi and state the value of your estimate.

(3)
题目中文翻译

一个半径为 1 cm 的圆盘被滚到由若干矩形组成的水平网格上，圆盘落在网格上任何位置的可能性相同。每个矩形长 5 cm、宽 3 cm；矩形之间没有空隙，且网格足够大，因此圆盘不会滚出网格。

若圆盘完全落在某个矩形内部且不覆盖该矩形的任何边，则赢得奖品。

通过考虑圆盘圆心可能落下的位置：

(a) 证明任意一次滚动赢得奖品的概率为 15\frac15。

第一组有 15 名学生，每人把圆盘滚到网格上 20 次，并记录各自获奖的次数 xx；结果由题中给出的 ∑x\sum x 与 ∑x2\sum x^2 概括。

(b) 求每名学生获奖次数的标准差。

第二组有 12 名学生，每人也滚动 20 次；每名学生获奖次数的平均数为 3.5，标准差为 2。

(c) 求全部 27 名学生每人获奖次数的平均数与标准差。

这 27 名学生还记录了圆盘覆盖矩形顶点的次数，并估计该事件的概率为 0.2216（精确到 4 位小数）。

(d) 说明如何使用这一概率估计 π\pi，并写出估计值。

解答

(a)

解法一

思路

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圆盘半径是 1 cm。若不碰到边，圆心必须离每条边至少 1 cm。因此在 5×35\times3 的矩形内，圆心可落的“安全区域”尺寸是 (5−2)×(3−2)=3×1(5-2)\times(3-2)=3\times1。

答题过程

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The centre of the disc must be at least 11 cm from each edge of the rectangle.

So the possible region for the centre is a rectangle of dimensions

(5−2) cm×(3−2) cm= 3 cm×1 cm.\begin{align*} (5-2)\text{ cm}\times(3-2)\text{ cm} =&\,3\text{ cm}\times1\text{ cm}. \end{align*}

The area of this region is

3×1=3.\begin{align*} 3\times1=3. \end{align*}

The area of one rectangle in the grid is

5×3=15.\begin{align*} 5\times3=15. \end{align*}

Since the centre is equally likely to land anywhere in the rectangle,

P(win)=315=15.\begin{align*} P(\text{win})=\frac{3}{15}=\frac15. \end{align*}

Hence the probability of winning a prize is 15\frac15, as required.

解法二

思路

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也可从总面积扣除圆心距离边界不足 1 cm 的区域。先扣上下两条 5×15\times1 带状区域，再扣左右两条剩余的 1×11\times1 区域，避免角落重复计算。

答题过程

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The total area of one grid rectangle is 5×3=155\times3=15.

The region within 11 cm of the top or bottom edge has area

2(5×1)=10.\begin{align*} 2(5\times1)=10. \end{align*}

After removing these two strips, the remaining regions within 11 cm of the left or right edge have total area

2(1×1)=2.\begin{align*} 2(1\times1)=2. \end{align*}

Therefore

P(win)= 1−10+215= 15.\begin{align*} P(\text{win}) =&\,1-\frac{10+2}{15}\\ =&\,\frac15. \end{align*}

Hence the required probability is 15\frac15.

(b)

解法一

思路

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用 σ=∑x2n−xˉ2\sigma=\sqrt{\frac{\sum x^2}{n}-\bar x^2}，其中 n=15n=15。

答题过程

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The mean for the first group is

xˉ=6115.\begin{align*} \bar x=\frac{61}{15}. \end{align*}

Therefore

σ= 29515−(6115)2= 1.7688….\begin{align*} \sigma =&\,\sqrt{\frac{295}{15}-\left(\frac{61}{15}\right)^2}\\ =&\,1.7688\ldots. \end{align*}

So the standard deviation is

1.77.\begin{align*} 1.77. \end{align*}

(c)

解法一

思路

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要合并两组数据，需要合并 ∑x\sum x 和 ∑x2\sum x^2。第二组已知平均数和标准差，可以反推出第二组的总和与平方和。

答题过程

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For the second group,

∑y=12(3.5)=42.\begin{align*} \sum y=12(3.5)=42. \end{align*}

Also,

22=∑y212−3.52.\begin{align*} 2^2=\frac{\sum y^2}{12}-3.5^2. \end{align*}

So

∑y2=12(22+3.52)=195.\begin{align*} \sum y^2=12(2^2+3.5^2)=195. \end{align*}

For the combined group,

∑z=61+42=103\begin{align*} \sum z=61+42=103 \end{align*}

and

∑z2=295+195=490.\begin{align*} \sum z^2=295+195=490. \end{align*}

The combined mean is

zˉ=10327=3.8148….\begin{align*} \bar z=\frac{103}{27}=3.8148\ldots. \end{align*}

The combined standard deviation is

σz= 49027−(10327)2= 1.8961….\begin{align*} \sigma_z =&\,\sqrt{\frac{490}{27}-\left(\frac{103}{27}\right)^2}\\ =&\,1.8961\ldots. \end{align*}

Therefore the mean and standard deviation are

3.81and1.90.\begin{align*} 3.81\quad\text{and}\quad1.90. \end{align*}

(d)

解法一

思路

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圆盘碰到某个角，等价于圆心落在该角周围半径 1 的圆形区域内。每个小矩形对应四个角的四分之一圆，总面积等于一个半径 1 的圆面积，即 π\pi。因此概率约为 π15\frac{\pi}{15}。

答题过程

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For the disc to cover a corner, the centre must be within 11 cm of a vertex.

Within one 5 cm×3 cm5\text{ cm}\times3\text{ cm} rectangle, the four possible corner regions combine to make one full circle of radius 11 cm.

So the probability is

π(1)25×3=π15.\begin{align*} \frac{\pi(1)^2}{5\times3}=\frac{\pi}{15}. \end{align*}

Using the estimate 0.22160.2216,

π15≈0.2216.\begin{align*} \frac{\pi}{15}\approx0.2216. \end{align*}

Therefore

π≈15(0.2216)=3.324.\begin{align*} \pi\approx15(0.2216)=3.324. \end{align*}