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IAL 2025 Jan Q7

A Level / Edexcel / S1

IAL 2025 Jan Paper · Question 7

题目

Problem

A farmer supplies cabbages to a local supermarket. The supermarket manager takes a random sample of 200200 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.

Box plot

(a) Use linear interpolation to estimate the probability that a randomly chosen cabbage from this sample has a weight of less than 570570 grams.

(2)

The supermarket manager takes the quartiles from this sample and now decides to classify any cabbage whose weight is at least Q3+1.5×(Q3−Q1)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.

(c) Explain whether or not the box plot supports this assumption.

(1)

The farmer records the weight of every cabbage harvested from a field and classifies any cabbage whose weight is more than 22 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 560560 grams. Given that a cabbage with a weight of 10001000 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)

解答

(a)

解法一

思路

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从 box plot 读出 median 是 550550，Q3Q_3 是 650650。570570 位于 median 和 Q3Q_3 之间，这一段包含上四分之一的数据。用线性插值估计 570570 在这一段中的比例。

答题过程

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From the box plot,

Q2=550,Q3=650.\begin{align*} Q_2=550,\qquad Q_3=650. \end{align*}

The probability below the median is 0.50.5, and the probability between Q2Q_2 and Q3Q_3 is 0.250.25.

Using linear interpolation between 550550 and 650650,

P(C<570)= 0.5+570−550650−550×0.25= 0.5+20100×0.25= 0.55.\begin{align*} P(C<570) =&\,0.5+\frac{570-550}{650-550}\times0.25\\[3mm] =&\,0.5+\frac{20}{100}\times0.25\\[3mm] =&\,0.55. \end{align*}

解法二

思路

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直观比例插值法。插值的本质是假设数据在区间内均匀分布。 从 Box plot 读出中位数 Q2=550Q_2 = 550，第三四分位数 Q3=650Q_3 = 650。 550550 以下的数据占 50%50\%。而区间 [550,650][550, 650] 的总长度为 100100，该区间内包含总数据的 25%25\%。 目标值 570570 占该区间的比例为：

570−550650−550=20100=20%\begin{align*} \frac{570-550}{650-550} = \frac{20}{100} = 20\% \end{align*}

因此，该区间内小于 570570 的数据占总数据的比例为：

20%×25%=5%=0.05\begin{align*} 20\% \times 25\% = 5\% = 0.05 \end{align*}

最后加上中位数下方的 50%50\%，即可得到总概率。这种方法避开了套用复杂的插值代数公式，完全依赖对分位数百分比物理意义的直观理解。

答题过程

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From the box plot, the median is Q2=550Q_2 = 550 and the upper quartile is Q3=650Q_3 = 650.

The data below 550550 accounts for 50%50\% of the distribution. The interval [550,650][550, 650] spans a width of 100100 grams and contains 25%25\% of the total data.

The position of 570570 within this interval is:

570−550650−550=20100=0.2.\begin{align*} \frac{570 - 550}{650 - 550} = \frac{20}{100} = 0.2. \end{align*}

Thus, the proportion of data between 550550 and 570570 is:

0.2×0.25=0.05.\begin{align*} 0.2 \times 0.25 = 0.05. \end{align*}

The total probability of cabbages weighing less than 570570 grams is:

P(C<570)=  0.50+0.05=  0.55.\begin{align*} P(C < 570) =&\,\, 0.50 + 0.05\\[3mm] =&\,\, 0.55. \end{align*}

(b)

解法一

思路

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从 box plot 读出 Q1=450Q_1=450、Q3=650Q_3=650，所以四分位距是 200200。代入题目给出的 extra-large 边界。

答题过程

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From the box plot,

Q1=450,Q3=650.\begin{align*} Q_1=450,\qquad Q_3=650. \end{align*}

So

Q3+1.5(Q3−Q1)= 650+1.5(650−450)= 650+1.5(200)= 950.\begin{align*} Q_3+1.5(Q_3-Q_1) =&\,650+1.5(650-450)\\[3mm] =&\,650+1.5(200)\\[3mm] =&\,950. \end{align*}

The minimum weight classified as extra-large is

950 grams.\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 Q1Q_1 到 median 和 median 到 Q3Q_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 22 standard deviations above the mean” 对应标准正态中的 Z>2Z>2。

答题过程

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For a normal distribution,

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

From the standard normal distribution,

P(Z>2)= 1−P(Z<2)= 1−0.9772= 0.0228.\begin{align*} P(Z>2) =&\,1-P(Z<2)\\[3mm] =&\,1-0.9772\\[3mm] =&\,0.0228. \end{align*}

(e)

解法一

思路

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Farmer 的 outlier 条件是重量超过 μ+2σ\mu+2\sigma。已知平均数是 560560，而 10001000 g 被归为 outlier，所以必须有

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

最大可能的 σ\sigma 来自临界值。

答题过程

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For 10001000 grams to be classified as an outlier,

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

So

440>2σ220>σ.\begin{align*} 440&>2\sigma\\[3mm] 220&>\sigma. \end{align*}

Hence the maximum possible value of the standard deviation is

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