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IAL 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)

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)\\[3mm] =&\,P(Z<1.5)\\[3mm] =&\,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)\\[3mm] =&\,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)\\[3mm] =&\,P\left(Z>\frac{-5}{10/1.96}\right)\\[3mm] =&\,P(Z>-0.98)\\[3mm] =&\,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)}\\[3mm] =&\,\frac{0.8365\ldots}{0.975}\\[3mm] =&\,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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比例对齐标准化法。我们并不需要真正求出标准差 σ\sigma。 由于 P(μ−10<T<μ+10)=0.95P(\mu-10 < T < \mu+10) = 0.95，对应的标准正态分布区间为 [−z10,z10][-z_{10}, z_{10}]。因为该区间的概率为 0.950.95，所以 z10=1.96z_{10} = 1.96。 这说明边界 μ−10\mu-10 对应标准正态分布的 ZZ 值为 −1.96-1.96。 由标准化的定义：

Z=T−μσ\begin{align*} Z = \frac{T - \mu}{\sigma} \end{align*}

由于 T=μ−5T = \mu - 5 偏离均值的距离为 −5-5，刚好是 μ−10\mu-10 偏离距离 −10-10 的一半，因此其对应的 ZZ 值也必然是 −1.96-1.96 的一半：

Z5=0.5×(−1.96)=−0.98\begin{align*} Z_5 = 0.5 \times (-1.96) = -0.98 \end{align*}

所以 P(T>μ−5)=P(Z>−0.98)P(T > \mu-5) = P(Z > -0.98)。利用这一比例线性关系，可以直接省去计算和写出 σ\sigma 的中间代数过程，极其高效且美妙。

答题过程

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From P(μ−10<T<μ+10)=0.95P(\mu-10 < T < \mu+10) = 0.95, the boundary T=μ−10T = \mu-10 corresponds to:

Z=−1.96.\begin{align*} Z = -1.96. \end{align*}

Since the difference between μ−5\mu - 5 and μ\mu is half of the difference between μ−10\mu - 10 and μ\mu, the standardized score for T=μ−5T = \mu - 5 is:

Z5=  0.5×(−1.96)=  −0.98.\begin{align*} Z_5 =&\,\, 0.5 \times (-1.96)\\[3mm] =&\,\, -0.98. \end{align*}

Thus, we have:

P(T>μ−5)=  P(Z>−0.98)=  P(Z<0.98)=  0.8365.\begin{align*} P(T > \mu - 5) =&\,\, P(Z > -0.98)\\[3mm] =&\,\, P(Z < 0.98)\\[3mm] =&\,\, 0.8365. \end{align*}

The conditional probability is:

P(T>μ−5∣T>μ−10)=  P(T>μ−5)P(T>μ−10)=  0.83650.975=  0.8579…\begin{align*} P(T > \mu-5 \mid T > \mu-10) =&\,\, \frac{P(T > \mu-5)}{P(T > \mu-10)}\\[3mm] =&\,\, \frac{0.8365}{0.975}\\[3mm] =&\,\, 0.8579\ldots \end{align*}

So the probability is 0.8580.858 (to 33 significant figures).