- My (wrong) process: I multiplied the median, instead of the mean, for the 2 years by the number of houses. Is that wrong because I assumed the house data to be evenly spaced?

The solution is (MEAN)*# of houses/total

Thank you!

- My (wrong) process: I multiplied the median, instead of the mean, for the 2 years by the number of houses. Is that wrong because I assumed the house data to be evenly spaced?

The solution is (MEAN)*# of houses/total

Thank you!

What option is correct?

You canâ€™t assume that the data is evenly spaced. If it was evenly spaced the mean and median would have been equal. You need to consider the Mean. Moreover, you can use the weighted average concept here to eliminate options C, D and E as in 2012 the number of houses were greater than that o 2013. So the average will be less than 275000.