CalcGospel 國際數學圖譜
返回

IAL 2020 Oct Q3

A Level / Edexcel / S1

IAL 2020 Oct Paper · Question 3

题目

Problem

The distance achieved in a long jump competition by students at a school is normally distributed with mean 3.8 metres and standard deviation 0.9 metres.

Students who achieve a distance greater than 4.3 metres receive a medal.

(a) Find the proportion of students who receive medals.

(3)

The school wishes to give a certificate of achievement or a medal to the 80% of students who achieve a distance of at least dd metres.

(b) Find the value of dd.

(3)

Of those who received medals, the 13\frac13 who jump the furthest will receive gold medals.

(c) Find the shortest distance, gg metres, that must be achieved to receive a gold medal.

(4)

A journalist from the local newspaper interviews a randomly selected group of 3 medal winners.

(d) Find the exact probability that there is at least one gold medal winner in the group.

(3)

解答

(a)

解法一

思路

展开

收 medal 表示 D>4.3D>4.3。标准化后查右尾概率。

答题过程

展开

Let DD be the distance jumped.

D∼N(3.8,0.92).\begin{align*} D\sim N(3.8,0.9^2). \end{align*}

Then

P(D>4.3)=P(Z>4.3−3.80.9)=P(Z>0.555…).P(D>4.3) =P\left(Z>\frac{4.3-3.8}{0.9}\right) =P(Z>0.555\ldots).

Using tables,

P(D>4.3)≈1−0.7123=0.2877.\begin{align*} P(D>4.3)\approx1-0.7123=0.2877. \end{align*}

So the proportion is approximately

0.288.\begin{align*} 0.288. \end{align*}

(b)

解法一

思路

展开

成绩至少 dd 的人占 80%，所以 dd 是左侧 20% 的分位点，对应 z=−0.8416z=-0.8416。

答题过程

展开

Since P(D≥d)=0.80P(D\geq d)=0.80,

P(D<d)=0.20.\begin{align*} P(D<d)=0.20. \end{align*}

Thus

d−3.80.9=−0.8416.\begin{align*} \frac{d-3.8}{0.9}=-0.8416. \end{align*}

So

d=3.8−0.8416(0.9)=3.0425….\begin{align*} d=3.8-0.8416(0.9)=3.0425\ldots. \end{align*}

Therefore

d=3.04 m.\begin{align*} d=3.04\text{ m}. \end{align*}

(c)

解法一

思路

展开

Gold 是 medal winners 中最远的三分之一，所以 P(D>g)P(D>g) 是 P(D>4.3)P(D>4.3) 的三分之一。

答题过程

展开

Since the top third of medal winners receive gold medals,

P(D>g)=13P(D>4.3).\begin{align*} P(D>g)=\frac13P(D>4.3). \end{align*}

Using part (a),

P(D>g)=13(0.289257…)=0.096419….\begin{align*} P(D>g)=\frac13(0.289257\ldots)=0.096419\ldots. \end{align*}

So

P(D<g)=1−0.096419…=0.903580….\begin{align*} P(D<g)=1-0.096419\ldots=0.903580\ldots. \end{align*}

This gives

g−3.80.9=1.3022….\begin{align*} \frac{g-3.8}{0.9}=1.3022\ldots. \end{align*}

Therefore

g=3.8+1.3022…(0.9)=4.972….\begin{align*} g=3.8+1.3022\ldots(0.9)=4.972\ldots. \end{align*}

So

g=4.97 m.\begin{align*} g=4.97\text{ m}. \end{align*}

(d)

解法一

思路

展开

在 medal winners 中，每个人是 gold winner 的概率是 13\frac13。至少一个通常用补集更快：1−1- 没有 gold 的概率。

答题过程

展开

For a medal winner,

P(gold)=13.\begin{align*} P(\text{gold})=\frac13. \end{align*}

Therefore

P(no gold)=(23)3.\begin{align*} P(\text{no gold})=\left(\frac23\right)^3. \end{align*}

Hence

P(at least one gold)=1−(23)3=1−827=1927.P(\text{at least one gold}) =1-\left(\frac23\right)^3 =1-\frac{8}{27} =\frac{19}{27}.