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IAL 2021 Oct Q3

A Level / Edexcel / S1

IAL 2021 Oct Paper · Question 3

题目

Problem

The stem and leaf diagram shows the ages of the 3535 male passengers on a cruise.

Stem and leaf diagram

(a) Find the median age of the male passengers.

(1)

(b) Show that the interquartile range (IQR) of these ages is 1616

(2)

An outlier is defined as a value that is more than 1.5×IQR⁡1.5\times\operatorname{IQR} above the upper quartile or 1.5×IQR⁡1.5\times\operatorname{IQR} below the lower quartile

(c) Show that there are 33 outliers amongst these ages.

(3)

(d) On the grid in Figure 1 on page 99, draw a box plot for the ages of the male passengers on the cruise.

Figure 1
(4)

Figure 1 on page 99 also shows a box plot for the ages of the female passengers on the cruise.

(e) Comment on any difference in the distributions of ages of male and female passengers on the cruise. State the values of any statistics you have used to support your comment.

(1)

Anja, along with her 22 daughters and a granddaughter, now join the cruise. Anja’s granddaughter is younger than both of Anja’s daughters. Anja had her 2323rd birthday on the day her eldest daughter was born. When their 44 ages are included with the other female passengers on the cruise, the box plot does not change.

(f) State, giving reasons, what you can say about

(i) the granddaughter’s age

(ii) Anja’s age.

(3)

解答

(a)

解法一

思路

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共有 3535 个数据，中位数是第 1818 个。

答题过程

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The median is the 1818th value, so

median=53.\begin{align*} \text{median}=53. \end{align*}

(b)

解法一

思路

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从 stem-and-leaf 读出 Q1=45Q_1=45，Q3=61Q_3=61，所以 IQR 为 1616。

答题过程

展开 Q1=45,Q3=61.\begin{align*} Q_1=45,\qquad Q_3=61. \end{align*}

Therefore

IQR⁡=61−45=16.\begin{align*} \operatorname{IQR}=61-45=16. \end{align*}

(c)

解法一

思路

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先算上下 outlier limits，再检查数据中哪些值越界。

答题过程

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The lower limit is

45−1.5(16)=21.\begin{align*} 45-1.5(16)=21. \end{align*}

The upper limit is

61+1.5(16)=85.\begin{align*} 61+1.5(16)=85. \end{align*}

The values outside these limits are

13, 87, 88.\begin{align*} 13,\ 87,\ 88. \end{align*}

Therefore there are 33 outliers.

(d)

解法一

思路

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箱体用 45,53,6145,53,61。由于 13,87,8813,87,88 是 outliers，whiskers 到非 outlier 的最小值和最大值：2727 与 7676。

答题过程

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Draw the male box plot using

lower whisker=27,Q1=45,Q2=53,Q3=61,upper whisker=76,outliers=13, 87, 88.\begin{gathered} \text{lower whisker}=27,\quad Q_1=45,\quad Q_2=53,\\[3mm] Q_3=61,\quad \text{upper whisker}=76,\\[3mm] \text{outliers}=13,\ 87,\ 88. \end{gathered}

(e)

解法一

思路

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比较时要点名统计量和数值。一个可行说法是女性乘客年龄整体更大，因为 female median 大于 male median。

答题过程

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The female passengers are generally older than the male passengers, because the median age for females is 6767, whereas the median age for males is 5353.

(f)

解法一

思路

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box plot 不变，说明新增的四个年龄不会改变已有的 quartiles、median、whiskers 或 outliers。孙女要落在女性 box plot 的下四分位以下但不改变下界；Anja 比大女儿大 2323 岁，并且也不能改变 box plot。

答题过程

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The granddaughter’s age must be between 3434 and 5656 inclusive, so that the lower part of the female box plot is unchanged.

Anja’s eldest daughter must be in the upper part of the female ages, and Anja is 2323 years older than her eldest daughter. Therefore Anja’s age must be between 9090 and 9393 inclusive.