题目

Problem

Students in class A and class B sit a statistics test. There are $24$ students in class A and $30$ students in class B.

The marks, $x$ , for students in each class are summarised in the table below.

	$n$	$\bar{x}$	$\sum x^2$
Class A	$24$	$47$	$66\,876$
Class B	$30$	$b$	$73\,826$

The two classes are combined into one group of $54$ students. The mean mark for all $54$ students is $45$ .

(a) Show that $b=43.4$

(2)

(b) Find the standard deviation of the marks for all $54$ students.

(2)

Following moderation, each student in class B has their mark increased by $2$ .

(i) the variance of the marks for class B

(ii) the mean mark for all $54$ students

(iii) the standard deviation of the marks for all $54$ students.

(3)

解答

(a)

解法一

思路

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总平均数是总分除以总人数。先用总平均数求出两班总分，再减去 A 班总分，得到 B 班总分，最后除以 $30$ 。

答题过程

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The total mark for all $54$ students is

\begin{align*} 54\times45=2430. \end{align*}

The total mark for class A is

\begin{align*} 24\times47=1128. \end{align*}

So the total mark for class B is

\begin{align*} 2430-1128=1302. \end{align*}

Therefore

\begin{align*} b =&\,\frac{1302}{30}\\[3mm] =&\,43.4. \end{align*}

Hence $b=43.4$ .

(b)

解法一

思路

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合并后的标准差用

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

这里 $\sum x^2$ 要把两个班的 $\sum x^2$ 相加， $\bar{x}=45$ ， $n=54$ 。

答题过程

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For all $54$ students,

\begin{align*} \sum x^2=66876+73826=140702. \end{align*}

Therefore

\begin{align*} s =&\,\sqrt{\frac{\sum x^2}{n}-\bar{x}^2}\\[3mm] =&\,\sqrt{\frac{140702}{54}-45^2}\\[3mm] =&\,\sqrt{2605.592\ldots-2025}\\[3mm] =&\,\sqrt{580.592\ldots}\\[3mm] =&\,24.095\ldots. \end{align*}

So the standard deviation is

\begin{align*} 24.1 \end{align*}

to $3$ significant figures.

解法二

思路

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方差合并加权公式。对于两组已知样本量、均值和方差的数据，其合并后的方差满足以下公式：

\begin{align*} s_{\text{overall}}^2 = \frac{n_A(s_A^2 + (\bar{x}_A - \bar{x}_{\text{overall}})^2) + n_B(s_B^2 + (\bar{x}_B - \bar{x}_{\text{overall}})^2)}{n_A + n_B} \end{align*}

我们可以先分别算出 Class A 和 Class B 的方差，然后利用该加权合并公式直接求出合并方差，最后开根号得到标准差。这种方法对于理解两组数据合并时“组内方差”与“组间方差”的贡献非常有教学价值。

答题过程

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First calculate the variance of each class.

For Class A:

\begin{align*} s_A^2 =&\,\, \frac{\sum x_A^2}{n_A} - \bar{x}_A^2\\[3mm] =&\,\, \frac{66876}{24} - 47^2\\[3mm] =&\,\, 2786.5 - 2209\\[3mm] =&\,\, 577.5 \end{align*}

For Class B:

\begin{align*} s_B^2 =&\,\, \frac{\sum x_B^2}{n_B} - \bar{x}_B^2\\[3mm] =&\,\, \frac{73826}{30} - 43.4^2\\[3mm] =&\,\, 2460.866\ldots - 1883.56\\[3mm] =&\,\, 577.306\ldots \end{align*}

Now use the weighted combined variance formula:

\begin{align*} s_{\text{overall}}^2 =&\,\, \frac{n_A(s_A^2 + (\bar{x}_A - \bar{x}_{\text{overall}})^2) + n_B(s_B^2 + (\bar{x}_B - \bar{x}_{\text{overall}})^2)}{n_A + n_B}\\[3mm] =&\,\, \frac{24\left(577.5 + (47 - 45)^2\right) + 30\left(577.306\ldots + (43.4 - 45)^2\right)}{54}\\[3mm] =&\,\, \frac{24(577.5 + 4) + 30(577.306\ldots + 2.56)}{54}\\[3mm] =&\,\, \frac{13956 + 17396}{54}\\[3mm] =&\,\, \frac{31352}{54}\\[3mm] =&\,\, 580.592\ldots \end{align*}

Finally, find the standard deviation:

\begin{align*} s_{\text{overall}} =&\,\, \sqrt{580.592\ldots}\\[3mm] =&\,\, 24.095\ldots \end{align*}

So the standard deviation is $24.1$ (to $3$ significant figures).

(c)

解法一

思路

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给 B 班每个人都加 $2$ ，不会改变 B 班内部的离散程度，所以 B 班方差不变。整体平均数一定增加，因为总分增加。整体标准差会减小，因为 B 班原本均值 $43.4$ 低于 A 班均值 $47$ ，加 $2$ 后变成 $45.4$ ，两班均值更接近。

答题过程

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(i) The variance of class B has no effect.

Adding $2$ to every mark in class B does not change the spread of the marks.

(ii) The mean mark for all $54$ students increases.

The total mark increases because each of the $30$ students in class B gains $2$ marks.

(iii) The standard deviation of the marks for all $54$ students decreases.

The mean mark of class B becomes closer to the mean mark of class A, so the marks are less spread out overall.

IAL 2025 May Q2

题目

解答

(a)

解法一

思路

答题过程

(b)

解法一

思路

答题过程

解法二

思路

答题过程

(c)

解法一

思路

答题过程