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IAL 2022 June Q3

A Level / Edexcel / S1

IAL 2022 June Paper · Question 3

题目

Problem

Gill buys a bag of logs to use in her stove. The lengths, ll cm, of the 8888 logs in the bag are summarised in the table below.

Length (l)Frequency (f)15<l⩽201920<l⩽253525<l⩽271627<l⩽301530<l⩽403\begin{array}{c|c} \text{Length }(l)&\text{Frequency }(f)\\ \hline 15<l\leqslant20&19\\ 20<l\leqslant25&35\\ 25<l\leqslant27&16\\ 27<l\leqslant30&15\\ 30<l\leqslant40&3 \end{array}

A histogram is drawn to represent these data. The bar representing logs with length 27<l⩽3027<l\leqslant30 has a width of 1.51.5 cm and a height of 44 cm.

(a) Calculate the width and height of the bar representing log lengths of 20<l⩽2520<l\leqslant25

(3)

(b) Use linear interpolation to estimate the median of ll

(2)

The maximum length of log Gill can use in her stove is 2626 cm. Gill estimates, using linear interpolation, that xx logs from the bag will fit into her stove.

(c) Show that x=62x=62

(1)

Gill randomly selects 44 logs from the bag.

(d) Using x=62x=62, find the probability that all 44 logs will fit into her stove.

(2)

The weights, WW grams, of the logs in the bag are coded using y=0.5w−255y=0.5w-255 and summarised by

n=88∑y=924∑y2=12862n=88\qquad \sum y=924\qquad \sum y^2=12862

(e) Calculate

(i) the mean of WW

(ii) the variance of WW

(6)

解答

(a)

解法一

思路

展开

真实 class width 33 cm 对应图上宽 1.51.5 cm，所以 55 cm 对应图上宽 2.52.5 cm。面积与频数成正比。

答题过程

展开

The scale for width is

3 cm real length⟶1.5 cm on histogram.\begin{align*} 3\text{ cm real length}\longrightarrow1.5\text{ cm on histogram}. \end{align*}

So for 20<l⩽2520<l\leqslant25, the histogram width is

2.5 cm.\begin{align*} 2.5\text{ cm}. \end{align*}

For 27<l⩽3027<l\leqslant30, the bar area is

1.5(4)=6.\begin{align*} 1.5(4)=6. \end{align*}

This represents 1515 logs, so the bar for 3535 logs has area

6⋅3515=14.\begin{align*} 6\cdot\frac{35}{15}=14. \end{align*}

Hence its height is

142.5=5.6 cm.\begin{align*} \frac{14}{2.5}=5.6\text{ cm}. \end{align*}

(b)

解法一

思路

展开

第 4444 个数据是中位数。累计到 2020 是 1919，所以中位数在 2020-2525 组内。

答题过程

展开 median= 20+44−1935(25−20)= 23.571…\begin{align*} \text{median} =&\,20+\frac{44-19}{35}(25-20)\\[3mm] =&\,23.571\ldots \end{align*}

So the median is

23.6 cm\begin{align*} 23.6\text{ cm} \end{align*}

to 33 significant figures.

(c)

解法一

思路

展开

2626 cm 包含前两整组 19+3519+35，再包含 2525 到 2626，也就是 2525-2727 组的一半。

答题过程

展开

The number of logs with l⩽25l\leqslant25 is

19+35=54.\begin{align*} 19+35=54. \end{align*}

Between 2525 and 2626 is half of the class 25<l⩽2725<l\leqslant27, so this contributes

12(16)=8.\begin{align*} \frac{1}{2}(16)=8. \end{align*}

Therefore

x=54+8=62.\begin{align*} x=54+8=62. \end{align*}

(d)

解法一

思路

展开

随机不放回抽 44 根，全部 fit 的概率是连续相乘。

答题过程

展开 P(all 4 fit)= 6288⋅6187⋅6086⋅5985= 0.23922…\begin{align*} P(\text{all 4 fit}) =&\,\frac{62}{88}\cdot\frac{61}{87} \cdot\frac{60}{86}\cdot\frac{59}{85}\\[3mm] =&\,0.23922\ldots \end{align*}

So the probability is

0.239\begin{align*} 0.239 \end{align*}

to 33 significant figures.

(e)(i)

解法一

思路

展开

由 y=0.5w−255y=0.5w-255 得 w=2(y+255)w=2(y+255)。先求 yˉ\bar{y}，再转回 wˉ\bar{w}。

答题过程

展开 yˉ=92488=10.5.\begin{align*} \bar{y}=\frac{924}{88}=10.5. \end{align*}

Since

w=2(y+255),\begin{align*} w=2(y+255), \end{align*}

we have

wˉ=2(10.5+255)=531.\begin{align*} \bar{w}=2(10.5+255)=531. \end{align*}

(e)(ii)

解法一

思路

展开

先求 yy 的 variance。因为 w=2y+510w=2y+510，variance 会乘以 22=42^2=4。

答题过程

展开 Var⁡(Y)= 1286288−10.52= 35.909…\begin{align*} \operatorname{Var}(Y) =&\,\frac{12862}{88}-10.5^2\\[3mm] =&\,35.909\ldots \end{align*}

Therefore

Var⁡(W)= 4Var⁡(Y)= 4(35.909…)= 143.636…\begin{align*} \operatorname{Var}(W) =&\,4\operatorname{Var}(Y)\\[3mm] =&\,4(35.909\ldots)\\[3mm] =&\,143.636\ldots \end{align*}

So

Var⁡(W)=144\begin{align*} \operatorname{Var}(W)=144 \end{align*}

to 33 significant figures.