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IAL 2021 Oct Q6

A Level / Edexcel / S1

IAL 2021 Oct Paper · Question 6

题目

Problem

Xiang is designing shelves for a bookshop. The height, HH cm, of books is modelled by the normal distribution with mean 25.125.1 cm and standard deviation 5.55.5 cm

(a) Show that P(H>30.8)=0.15P(H>30.8)=0.15

(3)

Xiang decided that the smallest 5%5\% of books and books taller than 30.830.8 cm would not be placed on the shelves. All the other books will be placed on the shelves.

(b) Find the range of heights of books that will be placed on the shelves.

(3)

The books that will be placed on the shelves have heights classified as small, medium or large. The numbers of small, medium and large books are in the ratios 2:3:32:3:3

(c) The medium books have heights xx cm where m<x<dm<x<d

(i) Show that d=25.8d=25.8 to 11 decimal place.

(3)

(ii) Find the value of mm

(4)

Xiang wants 22 shelves for small books, 33 shelves for medium books and 33 shelves for large books. These shelves will be placed one above another and made of wood that is 11 cm thick.

(d) Work out the minimum total height needed.

(2)

解答

(a)

解法一

思路

展开

标准化 30.830.8，再查右尾概率。

答题过程

展开 P(H>30.8)= P(Z>30.8−25.15.5)= P(Z>1.036…)= 1−0.8508= 0.1492…\begin{align*} P(H>30.8) =&\,P\left(Z>\frac{30.8-25.1}{5.5}\right)\\[3mm] =&\,P(Z>1.036\ldots)\\[3mm] =&\,1-0.8508\\[3mm] =&\,0.1492\ldots \end{align*}

Therefore

P(H>30.8)=0.15\begin{align*} P(H>30.8)=0.15 \end{align*}

to 22 significant figures.

(b)

解法一

思路

展开

最小的 5%5\% 不上架，所以下界是左侧概率 0.050.05 的分位数；上界是 30.830.8。

答题过程

展开

Let the lower boundary be yy.

Then

P(H<y)=0.05.\begin{align*} P(H<y)=0.05. \end{align*}

So

y−25.15.5=−1.6449.\begin{align*} \frac{y-25.1}{5.5}=-1.6449. \end{align*}

Hence

y=25.1−1.6449(5.5)=16.053…\begin{align*} y=25.1-1.6449(5.5)=16.053\ldots \end{align*}

Therefore the range of heights placed on the shelves is

16.1⩽H⩽30.8.\begin{align*} 16.1\leqslant H\leqslant30.8. \end{align*}

(c)(i)

解法一

思路

展开

上架的书占 80%80\%。比例 2:3:32:3:3 中，到 medium 上界 dd 为止，包括最小 5%5\%、small 的 20%20\%、medium 的 30%30\%，所以 P(H<d)=0.55P(H<d)=0.55。

答题过程

展开

For dd,

P(H<d)=0.05+0.20+0.30=0.55.\begin{align*} P(H<d)=0.05+0.20+0.30=0.55. \end{align*}

The corresponding standard normal value is approximately

z=0.13.\begin{align*} z=0.13. \end{align*}

So

d−25.15.5=0.13.\begin{align*} \frac{d-25.1}{5.5}=0.13. \end{align*}

Hence

d=25.1+0.13(5.5)=25.815.\begin{align*} d=25.1+0.13(5.5)=25.815. \end{align*}

Therefore

d=25.8\begin{align*} d=25.8 \end{align*}

to 11 decimal place.

(c)(ii)

解法一

思路

展开

mm 是 small 与 medium 的分界。到 mm 为止包括最小 5%5\% 和 small 的 20%20\%，所以 P(H<m)=0.25P(H<m)=0.25。

答题过程

展开

For mm,

P(H<m)=0.05+0.20=0.25.\begin{align*} P(H<m)=0.05+0.20=0.25. \end{align*}

The corresponding standard normal value is approximately

z=−0.6745.\begin{align*} z=-0.6745. \end{align*}

So

m−25.15.5=−0.6745.\begin{align*} \frac{m-25.1}{5.5}=-0.6745. \end{align*}

Hence

m=25.1−0.6745(5.5)=21.39…\begin{align*} m=25.1-0.6745(5.5)=21.39\ldots \end{align*}

Therefore

m=21.4.\begin{align*} m=21.4. \end{align*}

(d)

解法一

思路

展开

small shelves 需要高度 mm，medium shelves 需要高度 d=25.8d=25.8，large shelves 需要高度 30.830.8。共有 88 层木板厚度，每层 11 cm。

答题过程

展开

The minimum total height is

2(21.4)+3(25.8)+3(30.8)+8= 220.6.\begin{align*} 2(21.4)+3(25.8)+3(30.8)+8 =&\,220.6. \end{align*}

So the minimum total height needed is

221 cm\begin{align*} 221\text{ cm} \end{align*}

to the nearest cm.