CalcGospel 國際數學圖譜
返回

S1 2019 Jan Q3

A Level / Edexcel / S1

IAL 2019 Jan Paper · Question 3

题目

Problem

The weights of women boxers in a tournament are normally distributed with mean 64 kg and standard deviation 8 kg.

(a) Find the probability that a randomly chosen woman boxer in the tournament weighs less than 51 kg.

(3)

In the tournament, women boxers who weigh less than 51 kg are classified as lightweight. Ren weighs 49 kg and she has a match against another randomly selected, lightweight woman boxer.

(b) Find the probability that Ren weighs less than the other boxer.

(4)

In the tournament, women boxers who weigh more than HH kg are classified as heavyweight.

Given that 10% of the women boxers in the tournament are classified as heavyweight,

(c) find the value of HH.

(3)
题目中文翻译

某锦标赛中女子拳击手的体重服从平均数为 64 kg、标准差为 8 kg 的正态分布。

(a) 求随机选取的一名女子拳击手体重小于 51 kg 的概率。

体重小于 51 kg 的女子拳击手被划为轻量级。Ren 体重为 49 kg，她将与另一名随机选出的轻量级女子拳击手比赛。

(b) 求 Ren 的体重小于对手体重的概率。

体重大于 HH kg 的女子拳击手被划为重量级。已知锦标赛中 10% 的女子拳击手属于重量级。

(c) 求 HH。

解答

(a)

解法一

思路

展开

标准化 51，然后求左尾概率。

答题过程

展开

Let WW be the weight.

W∼N(64,82).\begin{align*} W\sim N(64,8^2). \end{align*}

Then

P(W<51)=P(Z<51−648)=P(Z<−1.625).P(W<51)=P\bigg(Z<\frac{51-64}{8}\bigg) =P(Z<-1.625).

Using tables,

P(W<51)≈0.052.\begin{align*} P(W<51)\approx0.052. \end{align*}

(b)

解法一

思路

展开

对手已知是 lightweight，所以条件是 W<51W<51。Ren weighs less than the other boxer 表示对手体重大于 49。

答题过程

展开

We need

P(W>49∣W<51).\begin{align*} P(W>49\mid W<51). \end{align*}

So

P(W>49∣W<51)=P(49<W<51)P(W<51).\begin{align*} P(W>49\mid W<51)=\frac{P(49<W<51)}{P(W<51)}. \end{align*}

Now

P(W<49)=P(Z<49−648)=P(Z<−1.875).P(W<49)=P\bigg(Z<\frac{49-64}{8}\bigg) =P(Z<-1.875).

Thus

P(49<W<51)=P(W<51)−P(W<49).\begin{align*} P(49<W<51)=P(W<51)-P(W<49). \end{align*}

Using the normal probabilities,

P(49<W<51)=0.02168….\begin{align*} P(49<W<51)=0.02168\ldots. \end{align*}

Therefore

P(W>49∣W<51)=0.02168…0.05208…=0.416….P(W>49\mid W<51) =\frac{0.02168\ldots}{0.05208\ldots} =0.416\ldots.

So the probability is approximately

0.42.\begin{align*} 0.42. \end{align*}

(c)

解法一

思路

展开

最重 10% 是右尾 0.10，对应 z=1.2816z=1.2816。

答题过程

展开

Since P(W>H)=0.10P(W>H)=0.10,

H−648=1.2816.\begin{align*} \frac{H-64}{8}=1.2816. \end{align*}

Therefore

H=64+1.2816(8)=74.2528.\begin{align*} H=64+1.2816(8)=74.2528. \end{align*}

So

H=74.3 kg.\begin{align*} H=74.3\text{ kg}. \end{align*}