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S1 2024 May Q5

A Level / Edexcel / S1

IAL 2024 May Paper · Question 5

题目

Problem

A competition consists of two rounds.

The time, in minutes, taken by adults to complete round one is modelled by a normal distribution with mean 1515 minutes and standard deviation 22 minutes.

(a) Use standardisation to find the proportion of adults that take less than 1818 minutes to complete round one.

(2)

Only the fastest 60%60\% of adults from round one take part in round two.

(b) Use standardisation to find the longest time that an adult can take to complete round one if they are to take part in round two.

(3)

The time, TT minutes, taken by adults to complete round two is modelled by a normal distribution with mean μ\mu.

Given that P(μ−10<T<μ+10)=0.95P(\mu-10<T<\mu+10)=0.95

(c) find P(T>μ−5∣T>μ−10)P(T>\mu-5\mid T>\mu-10).

(5)
题目中文翻译

一项比赛分为两轮。成年人完成第一轮所需的时间服从均值为 1515 min、标准差为 22 min 的正态分布。

(a) 利用标准化求成年人在少于 1818 min 内完成第一轮的比例。

只有第一轮中最快的 60%60\% 成年人可以参加第二轮。

(b) 利用标准化求能够参加第二轮的成年人完成第一轮所允许的最长时间。

成年人完成第二轮所需的时间 TT min 服从均值为 μ\mu 的正态分布。已知 P(μ−10<T<μ+10)=0.95P(\mu-10<T<\mu+10)=0.95。

(c) 求 P(T>μ−5∣T>μ−10)P(T>\mu-5\mid T>\mu-10)。

解答

(a)

解法一

思路

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设 XX 为 round one 用时，则 X∼N(15,22)X\sim N(15,2^2)。标准化 1818 后查表。

答题过程

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Let X∼N(15,22)X\sim N(15,2^2).

Then

P(X<18)= P(Z<18−152)= P(Z<1.5)= 0.9332.\begin{align*} P(X<18) =&\,P\left(Z<\frac{18-15}{2}\right)\\ =&\,P(Z<1.5)\\ =&\,0.9332. \end{align*}

Therefore the required proportion is

0.933.\begin{align*} 0.933. \end{align*}

(b)

解法一

思路

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最快的 60%60\% 能进入 round two，所以 cutoff time 是第 6060 百分位。标准正态表中左侧概率 0.60.6 对应 z≈0.2533z\approx0.2533。

答题过程

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Let the longest time be xx minutes. Since the fastest 60%60\% take part in round two,

P(X<x)=0.60.\begin{align*} P(X<x)=0.60. \end{align*}

The corresponding standard normal value is

z=0.2533.\begin{align*} z=0.2533. \end{align*}

So

x−152=0.2533.\begin{align*} \frac{x-15}{2}=0.2533. \end{align*}

Hence

x= 15+2(0.2533)= 15.5066.\begin{align*} x =&\,15+2(0.2533)\\ =&\,15.5066. \end{align*}

The longest time is approximately

15.5 minutes.\begin{align*} 15.5\text{ minutes}. \end{align*}

(c)

解法一

思路

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这是条件概率：

P(T>μ−5∣T>μ−10)=P(T>μ−5)P(T>μ−10).P(T>\mu-5\mid T>\mu-10) =\frac{P(T>\mu-5)}{P(T>\mu-10)}.

因为 T>μ−5T>\mu-5 一定包含在 T>μ−10T>\mu-10 里面，所以分子不用交集。

已知中间 μ±10\mu\pm10 的概率是 0.950.95，所以两侧尾部各 0.0250.025，因此 P(T>μ−10)=0.975P(T>\mu-10)=0.975，且 μ+10\mu+10 对应 z=1.96z=1.96。

答题过程

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Since

P(μ−10<T<μ+10)=0.95,\begin{align*} P(\mu-10<T<\mu+10)=0.95, \end{align*}

we have

P(T>μ−10)=0.975.\begin{align*} P(T>\mu-10)=0.975. \end{align*}

Also, μ+10\mu+10 is 1.961.96 standard deviations above the mean, so

10σ=1.96.\begin{align*} \frac{10}{\sigma}=1.96. \end{align*}

Thus

σ=101.96.\begin{align*} \sigma=\frac{10}{1.96}. \end{align*}

Now

P(T>μ−5)= P(Z>μ−5−μσ)= P(Z>−510/1.96)= P(Z>−0.98)= 0.8365….\begin{align*} P(T>\mu-5) =&\,P\left(Z>\frac{\mu-5-\mu}{\sigma}\right)\\ =&\,P\left(Z>\frac{-5}{10/1.96}\right)\\ =&\,P(Z>-0.98)\\ =&\,0.8365\ldots. \end{align*}

Therefore

P(T>μ−5∣T>μ−10)= P(T>μ−5)P(T>μ−10)= 0.8365…0.975= 0.8579….\begin{align*} P(T>\mu-5\mid T>\mu-10) =&\,\frac{P(T>\mu-5)}{P(T>\mu-10)}\\ =&\,\frac{0.8365\ldots}{0.975}\\ =&\,0.8579\ldots. \end{align*}

So

P(T>μ−5∣T>μ−10)=0.858\begin{align*} P(T>\mu-5\mid T>\mu-10)=0.858 \end{align*}

to 33 significant figures.

解法二

思路

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不必先求 σ\sigma。边界 μ−10\mu-10 对应 z=−1.96z=-1.96；μ−5\mu-5 与均值的距离是其一半，所以对应 z=−0.98z=-0.98。

答题过程

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The boundary T=μ−10T=\mu-10 corresponds to z=−1.96z=-1.96. Since 55 is half of 1010, the boundary T=μ−5T=\mu-5 corresponds to

z=−0.98.\begin{align*} z=-0.98. \end{align*} P(T>μ−5)= P(Z>−0.98)= 0.8365.\begin{align*} P(T>\mu-5) =&\,P(Z>-0.98)\\ =&\,0.8365. \end{align*} P(T>μ−5∣T>μ−10)= 0.83650.975= 0.8579…= 0.858(3 s.f.).\begin{align*} P(T>\mu-5\mid T>\mu-10) =&\,\frac{0.8365}{0.975}\\ =&\,0.8579\ldots\\ =&\,0.858\quad(3\text{ s.f.}). \end{align*}