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IAL 2024 Jan Q2

A Level / Edexcel / S1

IAL 2024 Jan Paper · Question 2

题目

Problem

The average minimum monthly temperature, xx degrees Fahrenheit (F)(^\circ\text{F}), and the average maximum monthly temperature, yy degrees Fahrenheit (F)(^\circ\text{F}), in Kolkata were recorded for 1212 months.

Some of the summary statistics are given below.

x=862,x2=62802,Syy=413.67,Sxy=512.67,n=12.\sum x=862,\quad \sum x^2=62802,\quad S_{yy}=413.67,\quad S_{xy}=512.67,\quad n=12.

(a) (i) Calculate the mean of the 1212 values of the average minimum monthly temperature.

(ii) Show that the standard deviation of the 1212 values of the average minimum monthly temperature is 8.57F8.57^\circ\text{F} to 33 significant figures.

(3)

(b) Calculate the product moment correlation coefficient between xx and yy.

(3)

For comparative purposes with a UK city, it was necessary to convert the temperatures from degrees Fahrenheit (F)(^\circ\text{F}) to degrees Celsius (C)(^\circ\text{C}). The formula used was

c=59(f32)c=\frac59(f-32)

where ff is the temperature in F^\circ\text{F} and cc is the temperature in C^\circ\text{C}.

(c) Use this formula and the values from part (a) to calculate, in C^\circ\text{C}, the mean and the standard deviation of the 1212 values of the average minimum monthly temperature in Kolkata. Give your answers to 33 significant figures.

(4)

Given that

  • uu is the equivalent temperature in C^\circ\text{C} of xx
  • vv is the equivalent temperature in C^\circ\text{C} of yy

(d) state, giving a reason, the product moment correlation coefficient between uu and vv.

(2)

解答

(a)(i)

解法一

思路

展开

平均数用 x/n\sum x/n

答题过程

展开 xˉ=86212=71.833.\begin{align*} \bar{x} =&\,\frac{862}{12}\\[3mm] =&\,71.833\ldots. \end{align*}

So the mean is

71.8F\begin{align*} 71.8^\circ\text{F} \end{align*}

to 33 significant figures.

(a)(ii)

解法一

思路

展开

标准差用

x2nxˉ2.\begin{align*} \sqrt{\frac{\sum x^2}{n}-\bar{x}^2}. \end{align*}

因为是 show that,要保留足够精度使用 xˉ=862/12\bar{x}=862/12

答题过程

展开 sd(X)=6280212(86212)2=73.4722=8.5715.\begin{align*} \operatorname{sd}(X) =&\,\sqrt{\frac{62802}{12}-\left(\frac{862}{12}\right)^2}\\[3mm] =&\,\sqrt{73.4722\ldots}\\[3mm] =&\,8.5715\ldots. \end{align*}

Therefore, to 33 significant figures,

sd(X)=8.57F.\begin{align*} \operatorname{sd}(X)=8.57^\circ\text{F}. \end{align*}

(b)

解法一

思路

展开

先求 SxxS_{xx},再用

r=SxySxxSyy.\begin{align*} r=\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}. \end{align*}

答题过程

展开 Sxx=x2(x)2n=62802862212=881.666.\begin{align*} S_{xx} =&\,\sum x^2-\frac{(\sum x)^2}{n}\\[3mm] =&\,62802-\frac{862^2}{12}\\[3mm] =&\,881.666\ldots. \end{align*}

Then

r=SxySxxSyy=512.67(881.666)(413.67)=0.8489.\begin{align*} r =&\,\frac{S_{xy}}{\sqrt{S_{xx}S_{yy}}}\\[3mm] =&\,\frac{512.67}{\sqrt{(881.666\ldots)(413.67)}}\\[3mm] =&\,0.8489\ldots. \end{align*}

So

r=0.849\begin{align*} r=0.849 \end{align*}

to 33 significant figures.

(c)

解法一

思路

展开

温度转换是线性变换:

c=59(f32).\begin{align*} c=\frac59(f-32). \end{align*}

平均数要先减 3232 再乘 59\frac59。标准差不受减 3232 影响,只会乘以 59\frac59

答题过程

展开

The mean in Celsius is

cˉ=59(xˉ32)=59(71.83332)=22.129.\begin{align*} \bar{c} =&\,\frac59(\bar{x}-32)\\[3mm] =&\,\frac59(71.833\ldots-32)\\[3mm] =&\,22.129\ldots. \end{align*}

So the mean is

22.1C.\begin{align*} 22.1^\circ\text{C}. \end{align*}

The standard deviation in Celsius is

59(8.5715)=4.7619.\begin{align*} \frac59(8.5715\ldots) =&\,4.7619\ldots. \end{align*}

So the standard deviation is

4.76C.\begin{align*} 4.76^\circ\text{C}. \end{align*}

(d)

解法一

思路

展开

PMCC 不受正斜率线性变换影响。华氏转摄氏对 x,yx,y 都是减常数再乘正数,所以相关系数不变。

答题过程

展开

The product moment correlation coefficient is still

0.849.\begin{align*} 0.849. \end{align*}

This is because adding or subtracting constants and multiplying by a positive constant do not change the product moment correlation coefficient.