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IAL 2024 Jan Q4

A Level / Edexcel / S1

IAL 2024 Jan Paper · Question 4

题目

Problem

A French test and a Spanish test were sat by 1111 students. The table below shows their marks.

StudentABCDEFGHIJK
French mark (f)(f)24243030323232323636363640404444505060606868
Spanish mark (s)(s)16169090242428283232363638384444484848486868

Greg says that if these points were plotted on a scatter diagram, then the point (30,90)(30,90) would be an outlier because 9090 is an outlier for the Spanish marks.

An outlier is defined as a value that is

greater than Q3+1.5(Q3Q1)\text{greater than }Q_3+1.5(Q_3-Q_1)

or

smaller than Q11.5(Q3Q1).\text{smaller than }Q_1-1.5(Q_3-Q_1).

(a) Show that 9090 is an outlier for the Spanish marks.

(3)

Ignoring the point (30,90)(30,90), Greg calculated the following summary statistics.

f=422,s=382,Sff=1667.6,Sfs=1735.6.\sum f=422,\qquad \sum s=382,\qquad S_{ff}=1667.6,\qquad S_{fs}=1735.6.

(b) Use these summary statistics to show that the equation of the least squares regression line of ss on ff for the remaining 1010 students is

s=5.72+1.04fs=-5.72+1.04f

where the values of the intercept and gradient are given to 33 significant figures. You must show your working.

(3)

(c) Give an interpretation of the gradient of the regression line.

(1)

Two further students sat the French test but missed the Spanish test.

(d) Using the equation given in part (b), estimate

(i) a Spanish mark for the student who scored 5555 marks in their French test,

(ii) a Spanish mark for the student who scored 1818 marks in their French test.

(3)

(e) State, giving a reason, which of the two estimates found in part (d) would be the more reliable estimate.

(2)

解答

(a)

解法一

思路

展开

把 Spanish marks 排序,找上下四分位数,再算上 outlier boundary。只要 9090 超过上界,就说明它是 outlier。

答题过程

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For the Spanish marks,

Q1=28,Q3=48.\begin{align*} Q_1=28,\qquad Q_3=48. \end{align*}

The upper outlier boundary is

Q3+1.5(Q3Q1)=48+1.5(4828)=48+30=78.\begin{align*} Q_3+1.5(Q_3-Q_1) =&\,48+1.5(48-28)\\[3mm] =&\,48+30\\[3mm] =&\,78. \end{align*}

Since

90>78,\begin{align*} 90>78, \end{align*}

9090 is an outlier for the Spanish marks.

(b)

解法一

思路

展开

回归线 s=a+bfs=a+bf 中,

b=SfsSff.\begin{align*} b=\frac{S_{fs}}{S_{ff}}. \end{align*}

然后用均值点求 aa

答题过程

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The gradient is

b=SfsSff=1735.61667.6=1.0407.\begin{align*} b =&\,\frac{S_{fs}}{S_{ff}}\\[3mm] =&\,\frac{1735.6}{1667.6}\\[3mm] =&\,1.0407\ldots. \end{align*}

For the remaining 1010 students,

fˉ=42210=42.2,sˉ=38210=38.2.\bar f=\frac{422}{10}=42.2,\qquad \bar s=\frac{382}{10}=38.2.

So

a=sˉbfˉ=38.2(1.0407)(42.2)=5.719.\begin{align*} a =&\,\bar s-b\bar f\\[3mm] =&\,38.2-(1.0407\ldots)(42.2)\\[3mm] =&\,-5.719\ldots. \end{align*}

Therefore, to 33 significant figures,

s=5.72+1.04f.\begin{align*} s=-5.72+1.04f. \end{align*}

(c)

解法一

思路

展开

斜率 1.041.04 表示 French mark 每增加 11 分,预测 Spanish mark 平均增加 1.041.04 分。

答题过程

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For each extra mark in the French test, the Spanish test mark is predicted to increase by about 1.041.04 marks.

(d)(i)

解法一

思路

展开

f=55f=55 代入回归方程。

答题过程

展开

When f=55f=55,

s=5.72+1.04(55)=51.48.\begin{align*} s =&\,-5.72+1.04(55)\\[3mm] =&\,51.48. \end{align*}

The estimate is

51.5.\begin{align*} 51.5. \end{align*}

(d)(ii)

解法一

思路

展开

f=18f=18 代入同一个回归方程。

答题过程

展开

When f=18f=18,

s=5.72+1.04(18)=13.0.\begin{align*} s =&\,-5.72+1.04(18)\\[3mm] =&\,13.0. \end{align*}

The estimate is

13.\begin{align*} 13. \end{align*}

(e)

解法一

思路

展开

剩余 1010 个学生的 French marks 范围是 242468685555 在范围内,是 interpolation;1818 在范围外,是 extrapolation。因此 5555 对应的估计更可靠。

答题过程

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The estimate for the student who scored 5555 in the French test is more reliable.

This is because 5555 is within the range of the French marks used to form the regression line, whereas 1818 is outside this range.