题目

Problem

A farmer supplies cabbages to a local supermarket. The supermarket manager takes a random sample of $200$ of these cabbages and the weight of each cabbage is recorded.

The weights of the cabbages in the sample are summarised in the following box plot.

(a) Use linear interpolation to estimate the probability that a randomly chosen cabbage from this sample has a weight of less than $570$ grams.

(2)

The supermarket manager takes the quartiles from this sample and now decides to classify any cabbage whose weight is at least $Q_3+1.5\times(Q_3-Q_1)$ as extra-large.

(b) Find the minimum weight of a cabbage that the supermarket manager would classify as extra-large.

(1)

The supermarket manager assumes that the weights of cabbages supplied to the supermarket may be modelled by a normal distribution.

(1)

The farmer records the weight of every cabbage harvested from a field and classifies any cabbage whose weight is more than $2$ standard deviations above the mean as an outlier. Assuming that the weights of the cabbages from this field may be modelled by a normal distribution,

(d) find the probability that a randomly selected cabbage harvested from the field would be classified as an outlier using the farmer’s method.

(2)

The mean weight of the cabbages harvested from the field is $560$ grams. Given that a cabbage with a weight of $1000$ grams is classified as an outlier by the farmer,

(e) find the maximum possible value of the standard deviation of the weights of cabbages harvested from the field.

(2)

题目中文翻译

一位农民向当地超市供应卷心菜。超市经理随机抽取 $200$ 个卷心菜，记录各自的重量。样本重量汇总在上方箱线图中。

(a) 使用线性插值，估计从样本中随机选取一个卷心菜，其重量小于 $570$ 克的概率。（2 分）

超市经理使用样本的四分位数，将重量至少为 $Q_3+1.5\times(Q_3-Q_1)$ 的卷心菜归为特大号。

(b) 求被归为特大号的卷心菜的最小重量。（1 分）

超市经理假设所供应的卷心菜重量可以用正态分布建模。

农民记录某块田地收获的每个卷心菜的重量，将重量高于均值超过 $2$ 个标准差的卷心菜归为离群值。假设这块田地的卷心菜重量可以用正态分布建模，

(d) 求从田地收获的卷心菜中随机选取一个，按农民方法被归为离群值的概率。（2 分）

这块田地收获的卷心菜平均重量为 $560$ 克。已知一个重 $1000$ 克的卷心菜被农民归为离群值，

(e) 求重量标准差的最大可能值。（2 分）

解答

(a)

解法一

思路

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从 box plot 读出 median 是 $550$ ， $Q_3$ 是 $650$ 。 $570$ 位于 median 和 $Q_3$ 之间，这一段包含上四分之一的数据。用线性插值估计 $570$ 在这一段中的比例。

答题过程

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From the box plot,

\begin{align*} Q_2=550,\qquad Q_3=650. \end{align*}

The probability below the median is $0.5$ , and the probability between $Q_2$ and $Q_3$ is $0.25$ .

Using linear interpolation between $550$ and $650$ ,

\begin{align*} P(C<570) =&\,0.5+\frac{570-550}{650-550}\times0.25\\ =&\,0.5+\frac{20}{100}\times0.25\\ =&\,0.55. \end{align*}

解法二

思路

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先用数量插值： $550$ 克对应第 $100$ 个， $650$ 克对应第 $150$ 个。求出 $570$ 克对应的数量，再除以总数 $200$ 。

答题过程

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Let $x$ be the estimated number of cabbages weighing less than $570$ grams. The cumulative frequencies at $550$ and $650$ grams are $100$ and $150$ .

\begin{align*} x=&\,100+(150-100)\frac{570-550}{650-550}\\ =&\,100+50\times\frac{20}{100}\\ x=&\,110. \end{align*}

Hence

\begin{align*} P(C<570)=&\,\frac{110}{200}=0.55. \end{align*}

(b)

解法一

思路

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从 box plot 读出 $Q_1=450$ 、 $Q_3=650$ ，所以四分位距是 $200$ 。代入题目给出的 extra-large 边界。

答题过程

展开

From the box plot,

\begin{align*} Q_1=450,\qquad Q_3=650. \end{align*}

\begin{align*} Q_3+1.5(Q_3-Q_1) =&\,650+1.5(650-450)\\ =&\,650+1.5(200)\\ =&\,950. \end{align*}

The minimum weight classified as extra-large is

\begin{align*} 950\text{ grams}. \end{align*}

(c)

解法一

思路

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Normal distribution 是对称的。box plot 看起来左右大致对称，尤其 $Q_1$ 到 median 和 median 到 $Q_3$ 距离相同，所以支持 normal model。

答题过程

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Yes, the box plot supports this assumption because it is reasonably symmetrical.

(d)

解法一

思路

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“More than $2$ standard deviations above the mean” 对应标准正态中的 $Z>2$ 。

答题过程

展开

For a normal distribution,

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

From the standard normal distribution,

\begin{align*} P(Z>2) =&\,1-P(Z<2)\\ =&\,1-0.9772\\ =&\,0.0228. \end{align*}

(e)

解法一

思路

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Farmer 的 outlier 条件是重量超过 $\mu+2\sigma$ 。已知平均数是 $560$ ，而 $1000$ g 被归为 outlier，所以必须有

\begin{align*} 1000>560+2\sigma. \end{align*}

最大可能的 $\sigma$ 来自临界值。

答题过程

展开

For $1000$ grams to be classified as an outlier,

\begin{align*} 1000>560+2\sigma. \end{align*}

\begin{align*} 440&>2\sigma\\ 220&>\sigma. \end{align*}

Hence the maximum possible value of the standard deviation is

\begin{align*} 220\text{ grams}. \end{align*}

S1 2025 Jan Q7

题目

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(a)

解法一

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解法二

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(b)

解法一

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(c)

解法一

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(d)

解法一

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(e)

解法一

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