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IAL 2020 Oct Q4

A Level / Edexcel / S1

IAL 2020 Oct Paper · Question 4

题目

Problem

A group of students took some tests. A teacher is analysing the average mark for each student. Each student obtained a different average mark.

For these average marks, the lower quartile is 24, the median is 30 and the interquartile range (IQR) is 10

The three lowest average marks are 8, 10 and 15.5 and the three highest average marks are 45, 52.5 and 56

The teacher defines an outlier to be a value that is either more than 1.5×IQR1.5\times\text{IQR} below the lower quartile or more than 1.5×IQR1.5\times\text{IQR} above the upper quartile

(a) Determine any outliers in these data.

(4)

(b) On the grid below draw a box plot for these data, indicating clearly any outliers.

(3)

(c) Use the quartiles to describe the skewness of these data. Give a reason for your answer.

(2)

Two more students also took the tests. Their average marks, which were both less than 45, are added to the data and the box plot redrawn.

The median and the upper quartile are the same but the lower quartile is now 26

(d) Redraw the box plot on the grid below.

(3)

(e) Give ranges of values within which each of these students’ average marks must lie.

(2)

解答

(a)

解法一

思路

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已知 Q1=24Q_1=24，IQR 为 10，所以 Q3=34Q_3=34。先算上下离群值边界，再判断给出的极端值。

答题过程

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Since

IQR⁡=Q3−Q1,\begin{align*} \operatorname{IQR}=Q_3-Q_1, \end{align*}

we have

Q3=24+10=34.\begin{align*} Q_3=24+10=34. \end{align*}

The lower outlier boundary is

24−1.5(10)=9.\begin{align*} 24-1.5(10)=9. \end{align*}

The upper outlier boundary is

34+1.5(10)=49.\begin{align*} 34+1.5(10)=49. \end{align*}

So outliers are values less than 99 or greater than 4949.

Therefore the outliers are

8,52.5,56.\begin{align*} 8,\quad 52.5,\quad 56. \end{align*}

(b)

解法一

思路

展开

盒子画 24,30,3424,30,34。非离群最小值是 10，非离群最大值是 45；8、52.5、56 要单独标成离群点。

答题过程

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Draw a box from 2424 to 3434, with the median at 3030.

Draw whiskers to 1010 and 4545.

Show 88, 52.552.5 and 5656 separately as outliers.

(c)

解法一

思路

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用中位数到上下四分位数的距离比较偏态。

答题过程

展开 Q2−Q1=30−24=6.\begin{align*} Q_2-Q_1=30-24=6. \end{align*}

Also,

Q3−Q2=34−30=4.\begin{align*} Q_3-Q_2=34-30=4. \end{align*}

Since

Q2−Q1>Q3−Q2,\begin{align*} Q_2-Q_1>Q_3-Q_2, \end{align*}

the data are negatively skewed.

(d)

解法一

思路

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新 Q1=26Q_1=26，中位数和 Q3Q_3 不变。重新计算 IQR 和离群值边界。

答题过程

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The new IQR is

34−26=8.\begin{align*} 34-26=8. \end{align*}

The new lower boundary is

26−1.5(8)=14.\begin{align*} 26-1.5(8)=14. \end{align*}

The new upper boundary is

34+1.5(8)=46.\begin{align*} 34+1.5(8)=46. \end{align*}

So 88 and 1010 are lower outliers, and 52.552.5 and 5656 are upper outliers.

The value 15.515.5 is not an outlier.

Draw a box from 2626 to 3434, with the median at 3030.

Draw whiskers to 15.515.5 and 4545.

Show 88, 1010, 52.552.5 and 5656 separately as outliers.

(e)

解法一

思路

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加入两个新数据后，Q1Q_1 从 24 变成 26，而中位数仍是 30，所以其中一个新数据必须在 26 到 30 之间；Q3=34Q_3=34 保持不变且两者都小于 45，所以另一个在 34 到 45 之间。

答题过程

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One of the new marks must lie in the range

26≤x≤30.\begin{align*} 26\leq x\leq30. \end{align*}

The other new mark must lie in the range

34≤x<45.\begin{align*} 34\leq x<45. \end{align*}