20 名女性顾客去买杂货。每位女性顾客所花费的时间（以分钟为单位）如茎叶图所示。

1. Use the modal time to identify the missing leaf in the 20s row

The modal time is $24$ minutes, so $24$ must occur more often than any other time.

In the stem $2$ row, the leaves are
$b,\ 3,\ c,\ 4,\ 4,\ 4,\ d.$
There are already three $24$s, while $18$ occurs three times in the stem $1$ row. Hence one more $24$ is needed.

So $d=4$.

2. Use the median and the two 23-minute shoppers to identify the remaining missing leaves in the 20s row

There are $20$ shoppers, so the median is the average of the $10$th and $11$th values.

There are $9$ values in the stem $1$ row, so the $10$th and $11$th values are the first two values in the stem $2$ row.

Since the median is $22$ minutes,
$\frac{20+b}{2}=22$
$20+b=44$
$b=24.$
As a leaf in the stem $2$ row, this means $b=2$.

Two shoppers spent $23$ minutes. One leaf $3$ is already shown in the stem $2$ row, so the other missing $23$ must be $c=3$.

3. Use ordering and range to identify the smallest and largest values

The leaves in each row are in ascending order. In the stem $1$ row,
$a,\ 0,\ 1,\ 4,\ 7,\ 8,\ 8,\ 8,\ 9$
so $a$ must be less than or equal to $0$. Since leaves are digits, $a=0$.

The smallest value is therefore $10$ minutes.

The range is $46$ minutes, so
$\text{largest value}-10=46$
$\text{largest value}=56.$
The largest value is in the stem $5$ row, so $e=6$.

1. Find the lower and upper quartiles for the female shoppers

From part (i), the ordered female times are
$10,10,11,14,17,18,18,18,19,22,23,23,24,24,24,24,32,32,35,56$.

For $20$ values,
$Q_1$ is the median of the first $10$ values:
$Q_1=\frac{17+18}{2}=17.5$
$Q_3$ is the median of the last $10$ values:
$Q_3=\frac{24+24}{2}=24$

2. Compare the female interquartile range with the male interquartile range

Female interquartile range:
$24-17.5=6.5\text{ minutes}$

Male interquartile range $=4.5$ minutes.

Since $6.5>4.5$, the female shoppers' times have a greater spread in the middle $50\%$ of the data.

1. Substitute the given values into the standard deviation formula

For $n=20$, $\sum x^2=9050$ and standard deviation $=4.53$,
$4.53=\sqrt{\frac{9050}{20}-\bar{x}^2}$

2. Solve the equation for the mean time

$4.53^2=\frac{9050}{20}-\bar{x}^2$
$20.5209=452.5-\bar{x}^2$
$\bar{x}^2=452.5-20.5209=431.9791$
$\bar{x}=\sqrt{431.9791}=20.784\ldots$
So the mean time is $20.8$ minutes, to 3 significant figures.