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S1 2024 Oct Q8

A Level / Edexcel / S1

IAL 2024 Oct Paper · Question 8

题目

Problem

An orchard produces apples.

The weights, AA grams, of its apples are normally distributed with mean μ\mu grams and standard deviation σ\sigma grams.

It is known that

P(A<162)=0.1andP(162<A<175)=0.7508.P(A<162)=0.1 \quad\text{and}\quad P(162<A<175)=0.7508.

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

(5)

A second orchard also produces apples.

The weights, BB grams, of its apples have distribution B∼N(215,102)B\sim N(215,10^2).

An outlier is a value that is

greater than Q3+1.5×(Q3−Q1)\text{greater than }Q_3+1.5\times(Q_3-Q_1)

or

smaller than Q1−1.5×(Q3−Q1).\text{smaller than }Q_1-1.5\times(Q_3-Q_1).

An apple is selected at random from this second orchard.

Using Q3=221.74Q_3=221.74 grams,

(b) find the probability that this apple is an outlier.

(5)
题目中文翻译

一个果园生产苹果。该果园苹果的重量 AA 克服从均值为 μ\mu 克、标准差为 σ\sigma 克的正态分布。

已知

P(A<162)=0.1且P(162<A<175)=0.7508.P(A<162)=0.1 \quad\text{且}\quad P(162<A<175)=0.7508.

(a) 计算 μ\mu 和 σ\sigma 的值。

另一个果园也生产苹果。该果园苹果的重量 BB 克服从分布 B∼N(215,102)B\sim N(215,10^2)。

离群值是满足下列条件之一的数值：

大于 Q3+1.5×(Q3−Q1)\text{大于 }Q_3+1.5\times(Q_3-Q_1)

或

小于 Q1−1.5×(Q3−Q1).\text{小于 }Q_1-1.5\times(Q_3-Q_1).

从第二个果园随机选取一个苹果。

使用 Q3=221.74Q_3=221.74 克，

(b) 求这个苹果为离群值的概率。

解答

(a)

解法一

思路

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P(A<162)=0.1P(A<162)=0.1 给出第 1010 百分位，对应标准正态值约为 −1.2816-1.2816。

又因为

P(162<A<175)=0.7508,\begin{align*} P(162<A<175)=0.7508, \end{align*}

所以

P(A<175)=0.8508.\begin{align*} P(A<175)=0.8508. \end{align*}

查表得 0.85080.8508 对应 z=1.04z=1.04。这就得到两个关于 μ,σ\mu,\sigma 的方程。

答题过程

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From

P(A<162)=0.1,\begin{align*} P(A<162)=0.1, \end{align*}

the corresponding zz-value is −1.2816-1.2816, so

162−μσ=−1.2816.\begin{align*} \frac{162-\mu}{\sigma}=-1.2816. \end{align*}

Hence

μ−1.2816σ=162.\begin{align*} \mu-1.2816\sigma=162. \end{align*}

Also,

P(A<175)= P(A<162)+P(162<A<175)= 0.1+0.7508= 0.8508.\begin{align*} P(A<175) =&\,P(A<162)+P(162<A<175)\\ =&\,0.1+0.7508\\ =&\,0.8508. \end{align*}

The corresponding zz-value is 1.041.04, so

175−μσ=1.04.\begin{align*} \frac{175-\mu}{\sigma}=1.04. \end{align*}

Hence

μ+1.04σ=175.\begin{align*} \mu+1.04\sigma=175. \end{align*}

Solving the two equations,

2.3216σ= 175−162= 13σ= 5.5995….\begin{align*} 2.3216\sigma=&\,175-162\\ =&\,13\\ \sigma=&\,5.5995\ldots. \end{align*}

Then

μ= 175−1.04(5.5995…)= 169.176….\begin{align*} \mu =&\,175-1.04(5.5995\ldots)\\ =&\,169.176\ldots. \end{align*}

Therefore

μ=169,σ=5.6\begin{align*} \mu=169,\qquad \sigma=5.6 \end{align*}

to 33 significant figures.

(b)

解法一

思路

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第二个 orchard 的分布关于平均数 215215 对称。已知

Q3=221.74,\begin{align*} Q_3=221.74, \end{align*}

所以

Q1=215−(221.74−215)=208.26.\begin{align*} Q_1=215-(221.74-215)=208.26. \end{align*}

先求 IQR，再求上、下离群边界。由于正态分布对称，两个尾部概率相同，可以求一边再乘以 22。

答题过程

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Since B∼N(215,102)B\sim N(215,10^2) is symmetric about 215215,

Q1= 215−(221.74−215)= 208.26.\begin{align*} Q_1 =&\,215-(221.74-215)\\ =&\,208.26. \end{align*}

So

Q3−Q1= 221.74−208.26= 13.48.\begin{align*} Q_3-Q_1 =&\,221.74-208.26\\ =&\,13.48. \end{align*}

The upper outlier limit is

221.74+1.5(13.48)= 241.96.\begin{align*} 221.74+1.5(13.48) =&\,241.96. \end{align*}

The lower outlier limit is

208.26−1.5(13.48)= 188.04.\begin{align*} 208.26-1.5(13.48) =&\,188.04. \end{align*}

Therefore

P(outlier)= P(B<188.04)+P(B>241.96).\begin{align*} P(\text{outlier}) =&\,P(B<188.04)+P(B>241.96). \end{align*}

By symmetry, these two probabilities are equal, so

P(outlier)= 2P(B>241.96)= 2P(Z>241.96−21510)= 2P(Z>2.696)≈ 2(0.0035)= 0.0070.\begin{align*} P(\text{outlier}) =&\,2P(B>241.96)\\ =&\,2P\Big(Z>\frac{241.96-215}{10}\Big)\\ =&\,2P(Z>2.696)\\ \approx&\,2(0.0035)\\ =&\,0.0070. \end{align*}

So the probability that the apple is an outlier is approximately

0.007.\begin{align*} 0.007. \end{align*}

解法二

思路

展开

先求苹果重量落在两个离群边界之间的概率，再用补集求离群概率。

答题过程

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The probability that the apple is not an outlier is

P(188.04<B<241.96)= P(−2.696<Z<2.696)= 0.9930 (approximately).\begin{align*} P(188.04<B<241.96) =&\,P(-2.696<Z<2.696)\\ =&\,0.9930\text{ (approximately)}. \end{align*}

Therefore

P(outlier)= 1−0.9930= 0.0070.\begin{align*} P(\text{outlier}) =&\,1-0.9930\\ =&\,0.0070. \end{align*}