2021年1月，21名女生参加了40米往返跑测试。成绩显示在茎叶图中。

1. Find the position of the median and read the corresponding value from the ordered stem-and-leaf diagram.

There are $21$ data values.

Median position $= \dfrac{21+1}{2}=11$th value.

Counting from the smallest value in the stem-and-leaf diagram, the 11th value is $11.2$.

So the median time is $11.2$ s.

1. Find the positions of the lower and upper quartiles for 21 data values.

There are $21$ timings.

The median is the $11$th value, so exclude the median and split the remaining values into two halves of $10$ values each.

$Q_1$ is the median of the first $10$ values, so it is the mean of the $5$th and $6$th values.

$Q_3$ is the median of the last $10$ values, so it is the mean of the $16$th and $17$th values.

2. Read the quartiles and calculate the interquartile range.

From the ordered stem-and-leaf diagram:

$5$th value $=10.6$, $6$th value $=10.7$

$Q_1=\frac{10.6+10.7}{2}=10.65$

$16$th value $=11.5$, $17$th value $=11.6$

$Q_3=\frac{11.5+11.6}{2}=11.55$

$\text{Interquartile range}=Q_3-Q_1=11.55-10.65=0.90$

1. Compare the median times to decide whether performance improved.

January median $=11.2$ s.

July median $=10.9$ s.

Since a shorter time means better performance and $10.9<11.2$, the performance improved in July.

2. Compare the interquartile ranges to decide whether the timings were more consistent.

January interquartile range $=1.1$ s.

For July,
$\text{IQR}=Q_3-Q_1=11.3-10.5=0.8\text{ s}.$

Since $0.8<1.1$, the July timings were less spread out and hence more consistent.