CalcGospel 國際數學圖譜
返回

IAL 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)

解答

(a)

解法一

思路

展开

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

展开

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

展开

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

展开

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\left(Z>\frac{31.2-30}{0.5}\right) =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*}