CalcGospel 國際數學圖譜
返回

S1 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)
题目中文翻译

某校学生参加跳远比赛，跳远距离服从均值为 3.8 米、标准差为 0.9 米的正态分布。

跳远距离超过 4.3 米的学生会获得奖牌。

(a) 求获得奖牌的学生比例。

学校希望向跳远距离至少为 dd 米的 80% 学生颁发成绩证书或奖牌。

(b) 求 dd 的值。

在获得奖牌的学生中，跳得最远的 13\frac13 将获得金牌。

(c) 求获得金牌所需的最短距离 gg 米。

当地报社的一名记者随机采访一组 3 名奖牌获得者。

(d) 求该组中至少有一名金牌获得者的精确概率。

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 the unrounded calculator value from 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}.