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IAL 2019 June Q2

A Level / Edexcel / S1

IAL 2019 June Paper · Question 2

题目

Problem

Chi wanted to summarise the scores of the 39 competitors in a village quiz. He started to produce the following stem and leaf diagram

Key: 2∣52\mid5 is a score of 25

He did not complete the stem and leaf diagram but instead produced the following box plot.

Chi defined an outlier as a value that is greater than Q3+1.5(Q3−Q1)Q_3+1.5(Q_3-Q_1) or less than Q1−1.5(Q3−Q1)Q_1-1.5(Q_3-Q_1)

(a) Find (i) the interquartile range (ii) the range.

(2)

(b) Describe, giving a reason, the skewness of the distribution of scores.

(2)

Albert and Beth asked for their scores to be checked.

Albert’s score was changed from 25 to 37

Beth’s score was changed from 54 to 60

(c) On the grid on page 5, draw an updated box plot. Show clearly any calculations that you used.

(7)

Some of the competitors complained that the questions were biased towards the younger generation. The product moment correlation coefficient between the age of the competitors and their score in the quiz is −0.187-0.187

(d) State, giving a reason, whether or not the complaint is supported by this statistic.

(2)

解答

(a)

解法一

思路

展开

从箱线图读出 Q1=33Q_1=33、Q3=47Q_3=47，最小值 11，最大值 54。

答题过程

展开 IQR⁡=47−33=14.\begin{align*} \operatorname{IQR}=47-33=14. \end{align*}

The range is

54−11=43.\begin{align*} 54-11=43. \end{align*}

(b)

解法一

思路

展开

比较中位数到上下四分位数的距离。左边距离 42−33=942-33=9，右边距离 47−42=547-42=5，左侧更长，所以负偏态。

答题过程

展开 Q2−Q1=42−33=9,\begin{align*} Q_2-Q_1=42-33=9, \end{align*}

whereas

Q3−Q2=47−42=5.\begin{align*} Q_3-Q_2=47-42=5. \end{align*}

Since Q2−Q1>Q3−Q2Q_2-Q_1>Q_3-Q_2, the distribution is negatively skewed.

(c)

解法一

思路

展开

改分后，下四分位数变成 35，上四分位数仍为 47，中位数仍为 42。重新计算 IQR 和离群值边界。

答题过程

展开

The new lower quartile is

Q1=35.\begin{align*} Q_1=35. \end{align*}

The median remains 4242, and the upper quartile remains 4747.

So the new IQR is

47−35=12.\begin{align*} 47-35=12. \end{align*}

The lower outlier boundary is

35−1.5(12)=17.\begin{align*} 35-1.5(12)=17. \end{align*}

The upper outlier boundary is

47+1.5(12)=65.\begin{align*} 47+1.5(12)=65. \end{align*}

So the outliers are 1111 and 1515.

The updated box plot should have a box from 3535 to 4747, median 4242, whiskers to 1818 and 6060, and outliers at 1111 and 1515.

(d)

解法一

思路

展开

−0.187-0.187 很接近 0，说明相关性很弱，不能支持“年龄越小分数越高”这种说法。

答题过程

展开

The complaint is not supported.

The product moment correlation coefficient −0.187-0.187 is close to 0, so there is only very weak correlation between age and score.