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CIE 9709 2025 November Paper 51 Q3

A Level / CIE / S1

CIE 9709 2025 Nov Paper 51 Paper · Question 3

题目

Problem

The back-to-back stem-and-leaf diagram shows the annual salaries, in dollars, of 27 employees at each of two companies, Browns and Greens.

(a) Find the median and interquartile range for the annual salaries of employees at Browns.

(3)

The annual salary of an employee at Browns is denoted by xx thousand dollars and the annual salary of an employee at Greens is denoted by yy thousand dollars. It is given that, for the 27 employees at each of the companies,

∑x=878.3,∑x2=28 616.09,\sum x=878.3,\qquad \sum x^2=28\,616.09, ∑y=896.5,∑y2=29 815.63.\sum y=896.5,\qquad \sum y^2=29\,815.63.

(b) Find the mean and standard deviation of the annual salaries of these 54 employees.

(5)
题目中文翻译

背靠背茎叶图显示了 Browns 和 Greens 两家公司各 27 名员工的年薪,单位为美元。

(a) 求 Browns 公司员工年薪的中位数和四分位距。

(3)

用 xx 千美元表示 Browns 公司一名员工的年薪,用 yy 千美元表示 Greens 公司一名员工的年薪。已知两家公司各自的 27 名员工满足

∑x=878.3,∑x2=28 616.09,\sum x=878.3,\qquad \sum x^2=28\,616.09, ∑y=896.5,∑y2=29 815.63.\sum y=896.5,\qquad \sum y^2=29\,815.63.

(b) 求这 54 名员工年薪的平均数和标准差。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

展开

共有27人,中位数为第14个值;去掉中位数后,两侧各13人,下、上四分位数为第7及第21个值。注意图中单位是百元,茎30表示三万元。

答题过程

展开

There are 27 salaries. Reading the ordered Browns data from the diagram:

  • the median is the 14th salary, 3240032400 dollars;
  • the lower quartile is the 7th salary, 3130031300 dollars;
  • the upper quartile is the 21st salary, 3350033500 dollars.

Hence the median is 3240032400 dollars and

interquartile range=33500−31300=2200 dollars.\text{interquartile range}=33500-31300=\boxed{2200\text{ dollars}}.

(b)

解法一

思路

展开

合并两组时,总和与平方和分别相加,人数也相加。不能直接平均两个标准差。数据单位是千元,最后均值与标准差都换回元。

答题过程

展开

In thousands of dollars, the combined total and sum of squares are

∑z=878.3+896.5=1774.8,\sum z=878.3+896.5=1774.8, ∑z2=28616.09+29815.63=58431.72.\sum z^2=28616.09+29815.63=58431.72.

For the 54 employees,

zˉ=1774.854=32.866666….\bar z=\frac{1774.8}{54}=32.866666\ldots.

Thus the mean salary is 32866.666…32866.666\ldots dollars, or 3290032900 dollars to 3 significant figures.

The combined variance, in squared thousands of dollars, is

σz2= 58431.7254−(1774.854)2= 1.851111….\begin{align*} \sigma_z^2=&\,\frac{58431.72}{54} -\biggl(\frac{1774.8}{54}\biggr)^2\\[4mm] =&\,1.851111\ldots. \end{align*}

Hence the standard deviation in dollars is

10001.851111…=1360.555…≈1360 dollars(3 s.f.).1000\sqrt{1.851111\ldots} =1360.555\ldots \approx\boxed{1360\text{ dollars}}\quad\text{(3 s.f.)}.