记录了一组 19 名学生在一次测试中的成绩，成绩如茎叶图所示。众数是 42 分。

1. Use the given mode to restrict the unknown leaf $a$.

The mark $42$ appears $3$ times.

Since the mode is $42$, the unknown mark $3a$ must not create another mark appearing $3$ times. In the stem 3 row, the leaves are
$a,\ 0,\ 1,\ 2,\ 6.$
For the leaves to remain in order, $a$ must be before $0$, so
$a=0.$

1. Determine the missing leaf from the ordered leaves in the stem 5 row.

In the stem $5$ row, the leaves are
$0,\ 2,\ 4,\ 4,\ b,\ 5,\ 8,\ 9$.

Since leaves are written in ascending order,
$4 \leq b \leq 5.$
As $b$ is a single-digit leaf and is placed before $5$ after two $4$s, the missing leaf is
$b=4.$

1. Locate the median position for 19 marks.

There are $19$ students.

Median position $=\dfrac{19+1}{2}=10$.

So the median is the $10$th mark in ascending order.

2. Read the 10th mark from the ordered stem-and-leaf diagram using the earlier values of $a$ and $b$.

Using $a=0$ and $b=4$, the marks in order are

$30,30,31,32,36,41,42,42,42,46,47,50,52,54,54,54,55,58,59$.

The $10$th mark is $42$.

1. Use all 19 marks to calculate the mean.

Using $a=0$ and $b=4$, the marks are
$30,30,31,32,36,41,42,42,42,46,47,50,52,54,54,54,55,58,59$.

$\text{Mean}=\frac{30+30+31+32+36+41+42+42+42+46+47+50+52+54+54+54+55+58+59}{19}$

$=\frac{855}{19}=45$

1. Use the completed list of marks and calculate the standard deviation from the mean.

The 19 marks are
$30,30,31,32,36,41,42,42,42,46,47,50,52,54,54,54,55,58,59.$
Using the mean from part (iii)(a), $\bar{x}=45$,
$\sigma=\sqrt{\frac{\sum (x-45)^2}{19}}.$
Now
$\sum (x-45)^2=1216.$
So
$\sigma=\sqrt{\frac{1216}{19}}=\sqrt{64}=8.$

1. Compare the average marks using the means.

For the group of 19 students, mean $=45$ marks.

For the group of 20 students, mean $=48.2$ marks.

Since $48.2>45$, the group of 20 students obtained higher marks on average.

2. Compare the spread or consistency of the marks using the standard deviations.

For the group of 19 students, standard deviation $=8.00$ marks.

For the group of 20 students, standard deviation $=7.22$ marks.

Since $7.22<8.00$, the group of 20 students had less spread in its marks, so its marks were more consistent.