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S1 2019 Oct Q6

A Level / Edexcel / S1

IAL 2019 Oct Paper · Question 6

题目

Problem

A machine cuts wood into pieces. The lengths, WW metres, of the pieces produced by the machine are normally distributed with mean μ\mu metres and standard deviation σ\sigma metres.

It is known that

P(W<3.968)=0.1andP(3.968<W<4.026)=0.75P(W<3.968)=0.1\quad\text{and}\quad P(3.968<W<4.026)=0.75

(a) Calculate the value of μ\mu and the value of σ\sigma

(5)

A second machine cuts wood into logs. The lengths, LL cm, of the logs produced by this second machine are normally distributed with L∼N(30,0.52)L\sim N(30,0.5^2)

An outlier is a value that is greater than Q3+1.5×(Q3−Q1)Q_3+1.5\times(Q_3-Q_1) or smaller than Q1−1.5×(Q3−Q1)Q_1-1.5\times(Q_3-Q_1)

A log is selected at random.

Given that Q1=29.7Q_1=29.7 to 3 significant figures,

(b) find the probability that the length of this log is an outlier.

(5)
题目中文翻译

一台机器把木材切成若干段。所得木段的长度 WW（米）服从平均数为 μ\mu 米、标准差为 σ\sigma 米的正态分布，并满足题目所给两个概率条件。

(a) 计算 μ\mu 与 σ\sigma。

另一台机器把木材切成原木。所得原木的长度 LL（厘米）服从 L∼N(30,0.52)L\sim N(30,0.5^2)。若一个数值大于 Q3+1.5(Q3−Q1)Q_3+1.5(Q_3-Q_1)，或小于 Q1−1.5(Q3−Q1)Q_1-1.5(Q_3-Q_1)，则定义为离群值。

随机选取一段原木。已知 Q1=29.7Q_1=29.7（3 位有效数字）。

(b) 求该原木长度为离群值的概率。

解答

(a)

解法一

思路

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把两个概率条件翻译成标准正态分位点。左侧 10% 对应 z=−1.2816z=-1.2816，从 3.968 到 4.026 的概率是 0.75，所以 P(W<4.026)=0.85P(W<4.026)=0.85，对应 z=1.0364z=1.0364。

答题过程

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Since

P(W<3.968)=0.1,\begin{align*} P(W<3.968)=0.1, \end{align*}

we have

3.968−μσ=−1.2816.\begin{align*} \frac{3.968-\mu}{\sigma}=-1.2816. \end{align*}

So

μ−1.2816σ=3.968.\begin{align*} \mu-1.2816\sigma=3.968. \end{align*}

Also,

P(W<4.026)=0.1+0.75=0.85.\begin{align*} P(W<4.026)=0.1+0.75=0.85. \end{align*}

Thus

4.026−μσ=1.0364,\begin{align*} \frac{4.026-\mu}{\sigma}=1.0364, \end{align*}

so

μ+1.0364σ=4.026.\begin{align*} \mu+1.0364\sigma=4.026. \end{align*}

Subtracting the two equations,

2.3180σ=0.058.\begin{align*} 2.3180\sigma=0.058. \end{align*}

Therefore

σ=0.0250….\begin{align*} \sigma=0.0250\ldots. \end{align*}

Using μ+1.0364σ=4.026\mu+1.0364\sigma=4.026,

μ=4.026−1.0364(0.0250…)=4.000….\begin{align*} \mu=4.026-1.0364(0.0250\ldots)=4.000\ldots. \end{align*}

Hence

μ=4.00,σ=0.025.\begin{align*} \mu=4.00,\qquad \sigma=0.025. \end{align*}

(b)

解法一

思路

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正态分布关于均值 30 对称，所以 Q3Q_3 与 Q1Q_1 对称，约为 30.3。先算 outlier limits，再求两侧尾部概率。

答题过程

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Since the distribution is symmetric about 3030,

Q3≈30.3.\begin{align*} Q_3\approx30.3. \end{align*}

So

IQR⁡=30.3−29.7=0.6.\begin{align*} \operatorname{IQR}=30.3-29.7=0.6. \end{align*}

The outlier limits are

29.7−1.5(0.6)=28.8\begin{align*} 29.7-1.5(0.6)=28.8 \end{align*}

and

30.3+1.5(0.6)=31.2.\begin{align*} 30.3+1.5(0.6)=31.2. \end{align*}

Thus

P(outlier)=P(L<28.8)+P(L>31.2).\begin{align*} P(\text{outlier})=P(L<28.8)+P(L>31.2). \end{align*}

By symmetry,

P(outlier)=2P(L>31.2).\begin{align*} P(\text{outlier})=2P(L>31.2). \end{align*}

Now

P(L>31.2)=P(Z>31.2−300.5)=P(Z>2.4).P(L>31.2)=P\bigg(Z>\frac{31.2-30}{0.5}\bigg) =P(Z>2.4).

So

P(L>31.2)=0.0082.\begin{align*} P(L>31.2)=0.0082. \end{align*}

Therefore

P(outlier)=2(0.0082)=0.0164.\begin{align*} P(\text{outlier})=2(0.0082)=0.0164. \end{align*}