题目

Problem

An orchard produces apples.

The weights, $A$ grams, of its apples are normally distributed with mean $\mu$ grams and standard deviation $\sigma$ grams.

It is known that

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

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

(5)

A second orchard also produces apples.

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

An outlier is a value that is

\text{greater than }Q_3+1.5(Q_3-Q_1)

\text{smaller than }Q_1-1.5(Q_3-Q_1).

An apple is selected at random from this second orchard.

Using $Q_3=221.74$ grams,

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

(5)

解答

(a)

解法一

思路

展开

$P(A<162)=0.1$ 给出第 $10$ 百分位，对应标准正态值约为 $-1.2816$ 。

又因为

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

所以

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

查表得 $0.8508$ 对应 $z=1.04$ 。这就得到两个关于 $\mu,\sigma$ 的方程。

答题过程

展开

From

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

the corresponding $z$ -value is $-1.2816$ , so

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

Hence

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

Also,

\begin{align*} P(A<175) =&\,P(A<162)+P(162<A<175)\\[3mm] =&\,0.1+0.7508\\[3mm] =&\,0.8508. \end{align*}

The corresponding $z$ -value is $1.04$ , so

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

Hence

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

Solving the two equations,

\begin{align*} (\mu+1.04\sigma)-(\mu-1.2816\sigma) =&\,175-162\\[3mm] 2.3216\sigma=&\,13\\[3mm] \sigma=&\,5.5995\ldots. \end{align*}

Then

\begin{align*} \mu =&\,175-1.04(5.5995\ldots)\\[3mm] =&\,169.176\ldots. \end{align*}

Therefore

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

to $3$ significant figures.

(b)

解法一

思路

展开

第二个 orchard 的分布关于平均数 $215$ 对称。已知

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

所以

\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 $2$ 。

答题过程

展开

Since $B\sim N(215,10^2)$ is symmetric about $215$ ,

\begin{align*} Q_1 =&\,215-(221.74-215)\\[3mm] =&\,208.26. \end{align*}

\begin{align*} Q_3-Q_1 =&\,221.74-208.26\\[3mm] =&\,13.48. \end{align*}

The upper outlier limit is

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

The lower outlier limit is

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

Therefore

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

By symmetry, these two probabilities are equal, so

\begin{align*} P(\text{outlier}) =&\,2P(B>241.96)\\[3mm] =&\,2P\left(Z>\frac{241.96-215}{10}\right)\\[3mm] =&\,2P(Z>2.696)\\[3mm] &\approx2(0.0035)\\[3mm] =&\,0.0070. \end{align*}

So the probability that the apple is an outlier is approximately

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

解法二

思路

展开

分立概率求和法。不直接利用对称性，而是分别计算出下离群（小于 $188.04$ ）和上离群（大于 $241.96$ ）的概率，然后把它们加起来。这样可以更直白地验证两个尾部的概率大小，也适合对正态分布对称性应用不太熟练的同学。

答题过程

展开

Calculate the lower outlier probability:

\begin{align*} P(B < 188.04) =&\,\, P\left(Z < \frac{188.04 - 215}{10}\right)\\[3mm] =&\,\, P(Z < -2.696)\\[3mm] =&\,\, 1 - P(Z < 2.696)\\[3mm] \approx&\,\, 1 - 0.9965\\[3mm] =&\,\, 0.0035. \end{align*}

Calculate the upper outlier probability:

\begin{align*} P(B > 241.96) =&\,\, P\left(Z > \frac{241.96 - 215}{10}\right)\\[3mm] =&\,\, P(Z > 2.696)\\[3mm] =&\,\, 1 - P(Z < 2.696)\\[3mm] \approx&\,\, 1 - 0.9965\\[3mm] =&\,\, 0.0035. \end{align*}

Sum the two outlier probabilities:

\begin{align*} P(\text{outlier}) =&\,\, P(B < 188.04) + P(B > 241.96)\\[3mm] =&\,\, 0.0035 + 0.0035\\[3mm] =&\,\, 0.0070. \end{align*}

So the probability that the apple is an outlier is approximately $0.007$ .

IAL 2024 Oct Q8

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程