茎叶图显示了某食品店分店 A 在一个月内的每日顾客人数。

((i))
写出分店 A 顾客人数的中位数。

答案 -------------------------------
((ii))
求分店 A 的四分位距。

答案 -------------------------------
((iii))
另一家分店（分店 B）的顾客人数中位数为 27，四分位距为 23。该食品店打算仅扩张这两家分店中的一家。应该扩张哪一家分店？清晰地解释你的选择。

答案 ---------------------------------------------------------------------------------------------------------------------------------

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完整解法

1. Identify the middle positions and read the corresponding values from the ordered stem-and-leaf diagram.

There are $30$ values in the stem-and-leaf diagram.

For an even number of values, the median is the average of the $15$ th and $16$ th values.

From the ordered diagram, the $15$ th value is $33$ and the $16$ th value is $34$ .

$\text{Median}=\frac{33+34}{2}=33.5$

答案

$33.5$ customers

1. Identify the positions of the lower and upper quartiles in the ordered data.

There are $30$ values in the stem-and-leaf diagram.

Since $30$ is even, split the ordered data into two halves of $15$ values each.

$Q_1$ is the median of the lower $15$ values, so it is the $8$ th value.

$Q_3$ is the median of the upper $15$ values, so it is the $23$ rd value overall.

2. Read the quartiles and calculate the interquartile range.

From the stem-and-leaf diagram,

$Q_1 = 22$ and $Q_3 = 47$ .

$\text{Interquartile range} = Q_3 - Q_1 = 47 - 22 = 25$

答案

$25$ customers

1. Compare the medians and justify the choice of branch.

Branch A: median $=33.5$ , interquartile range $=25$ .

Branch B: median $=27$ , interquartile range $=23$ .

Since Branch A has a higher median number of customers, it has a higher typical daily number of customers. Hence Branch A should be expanded.

答案

Branch A should be expanded, since it has a higher median number of customers, $33.5$ compared with $27$ for Branch B.

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