I am going through the “Statistics problems” section and found that the answers to the below question made sense. However, when going to the review section, it seems like only the mean average matters as the median is discarded. Can anyone please provide clarity on why this is?
The following table indicates the number of paying customers at a local restaurant by week.
|Week number|Number of customers|
For what values of x are both the mean and the median of paying customers for the entire six-week period between 70 and 80?
Indicate all that apply.
FYI: I am referring to the solution offered for CYY64-RBG6D
Hi @Ana2, welcome to Achievable!
The trick here is in the list of numbers, and which numbers get used in the median calculation:
Textbook version: 13, 23, 67, 100, 198 (median: 67)
Quiz version: 50, 65, 65, 71, 90 (median: 65)
Let’s think about what would happen if we added in the number 31.
Textbook version: 13, 23, 31, 67, 100, 198 (median changed: 49)
Quiz version: 31, 50, 65, 65, 71, 90 (median unchanged: 65)
So basically, for the quiz version, any number that is <=65 or >=71 will not change the median. Since the current list’s median falls within the range already, we can essentially ignore that condition.