累积频数曲线显示了 180 名学生在 2021 年一次数学考试中的成绩分布。最高可能分数为 80 分。

1. Find the number of students who scored less than or equal to $n$ marks.

Since $\dfrac{5}{6}$ of the students scored more than $n$ marks, the fraction who scored less than or equal to $n$ marks is

$1-\frac{5}{6}=\frac{1}{6}.$

Number of students scoring $\leq n$ marks:

$\frac{1}{6}\times 180=30.$

2. Read the mark corresponding to cumulative frequency 30 from the ogive.

On the cumulative frequency curve, draw a horizontal line from cumulative frequency $30$ to the curve, then draw a vertical line down to the marks axis.

From the graph,

$n \approx 35.$

1. Find the original examination mark corresponding to a final mark of $37$.

Let the original mark be $x$.

Final mark $= \dfrac{x}{2}+15$.

For a final mark of $37$:

$\dfrac{x}{2}+15=37$

$\dfrac{x}{2}=22$

$x=44$

So students with final mark $37$ or less had original marks $44$ or less.

2. Read the cumulative frequency for marks less than or equal to $44$ from the ogive.

From the cumulative frequency curve, at $44$ marks, the cumulative frequency is approximately

$82.$

Hence, the number of students with final mark $37$ or less is approximately $82$.

1. Relate the same median to the cumulative frequency curve.

There are $180$ students, so the median is at cumulative frequency
$\frac{180}{2}=90.$
Since the new group has the same median, its cumulative frequency curve passes through the same mark at cumulative frequency $90$.

2. Describe the effect of a larger standard deviation on the curve.

A larger standard deviation means the marks are more spread out from the median. Hence the cumulative frequency curve for the new group will be more stretched horizontally and less steep around the middle, while keeping the same median point.

1. Find the number of students in each mark category.

From the cumulative frequency curve, the number of students who scored $60$ marks or less is approximately $170$.

Total number of students $=180$.

Number who scored more than $60$ marks:
$180-170=10$

2. Find the probability of choosing one student from each category.

The two students are chosen without replacement.

Required probability:
$\frac{170}{180}\times\frac{10}{179}+\frac{10}{180}\times\frac{170}{179}$
$=2\left(\frac{170}{180}\times\frac{10}{179}\right)$
$=\frac{3400}{32220}$
$=\frac{170}{1611}$