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IAL 2023 Oct Q2

A Level / Edexcel / S1

IAL 2023 Oct Paper · Question 2

题目

Problem

The weights, to the nearest kilogram, of a sample of 3333 red kangaroos taken in December are summarised in the stem and leaf diagram below.

Stem and leaf diagram

(a) Find

(i) the value of the median

(ii) the value of Q1Q_1 and the value of Q3Q_3

for the weights of these red kangaroos.

(3)

For these data an outlier is defined as 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)

(b) Show that there are 22 outliers for these data.

(3)

Figure 1 on page 77 shows a box plot for the weights of the same 3333 red kangaroos taken in February, earlier in the year.

(c) In the space on Figure 1, draw a box plot to represent the weights of these red kangaroos in December.

Figure 1
(4)

(d) Compare the distribution of the weights of red kangaroos taken in February with the distribution of the weights of red kangaroos taken in December of the same year. You should interpret your comparisons in the context of the question.

(3)

解答

(a)

解法一

思路

展开

共有 3333 个数据，所以中位数是第 1717 个。下四分位数是前 1616 个数据的中位数，即第 88 和第 99 个的平均；上四分位数是后 1616 个数据的中位数，即第 2525 和第 2626 个的平均。

答题过程

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There are 3333 values.

The median is the 1717th value, so

Q2=57.\begin{align*} Q_2=57. \end{align*}

For the lower quartile,

Q1=8th value+9th value2=45+452=45.Q_1=\frac{\text{8th value}+\text{9th value}}{2} =\frac{45+45}{2}=45.

For the upper quartile,

Q3=25th value+26th value2=63+632=63.Q_3=\frac{\text{25th value}+\text{26th value}}{2} =\frac{63+63}{2}=63.

(b)

解法一

思路

展开

先算四分位距 Q3−Q1Q_3-Q_1，再算上下界。比下界小或比上界大的数据就是 outlier。

答题过程

展开 IQR⁡= Q3−Q1= 63−45= 18.\begin{align*} \operatorname{IQR} =&\,Q_3-Q_1\\[3mm] =&\,63-45\\[3mm] =&\,18. \end{align*}

The upper outlier boundary is

Q3+1.5(Q3−Q1)=63+1.5(18)=90.\begin{align*} Q_3+1.5(Q_3-Q_1)=63+1.5(18)=90. \end{align*}

The lower outlier boundary is

Q1−1.5(Q3−Q1)=45−1.5(18)=18.\begin{align*} Q_1-1.5(Q_3-Q_1)=45-1.5(18)=18. \end{align*}

The values outside these boundaries are

16and94.\begin{align*} 16 \quad\text{and}\quad 94. \end{align*}

Therefore there are 22 outliers.

(c)

解法一

思路

展开

画 December 的 box plot 时，箱体用 Q1=45Q_1=45、median =57=57、Q3=63Q_3=63。由于 1616 和 9494 是 outliers，须单独标出；须把 whiskers 画到非 outlier 的最小值和最大值，也就是 2323 与 8686。

答题过程

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For December, draw the box plot using:

Q1=45,Q2=57,Q3=63,lower whisker=23,upper whisker=86,outliers=16, 94.\begin{gathered} Q_1=45,\qquad Q_2=57,\qquad Q_3=63,\\[3mm] \text{lower whisker}=23,\qquad \text{upper whisker}=86,\\[3mm] \text{outliers}=16,\ 94. \end{gathered}

(d)

解法一

思路

展开

比较两个 box plot，至少要比较一个平均水平和一个离散程度，并且要放回红袋鼠体重的语境中说。

答题过程

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The median weight in February is lower than the median weight in December. This suggests that, on average, the red kangaroos weighed less in February than in December.

The interquartile range in February is lower than the interquartile range in December. This suggests that the weights of the red kangaroos were less varied in February.