Answer:
(a)
Explanation:
Given
See attachment for sets A and B
Required
The true statement about both sets
First, we calculate the typical values (mean) of set A and set B.
This is calculated as:

For A:



For B:



Here, we can conclude that B has a larger typical value
Next calculate the spread (range) of sets A and B
This is calculated as:

For A:


For B


Here, we can conclude that A has a larger spread.
Hence, (a) is true