This community-built FAQ covers the “Square The Differences” exercise from the lesson “Variance”.
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This exercise can be found in the following Codecademy content:
Learn Statistics With Python
FAQs on the exercise Square The Differences
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1 Like
Hello guys,
I we simply want to get rid of the negatives, why not take the absolute value instead of the squared value?
Intuitively that seems more accurate.
1 Like
I had the same doubt! This page here explains “Why Square?”: Standard Deviation and Variance (mathsisfun.com). In short, using absolute values (which results in calculating the “Mean Deviation” for the dataset) wouldn’t truly account for the difference in datasets such as these two:
set1 = [4, 4, -4, -4] # Less spread out
np.mean(set1) # Equals 0
absolute_values = [4, 4, 4, 4]
mean_deviation = (4 + 4 + 4 + 4) / 4 # Equals 4
set2 = [7, 1, -6, -2] # More spread out
np.mean(set2) # Equals 0
absolute_values = [7, 1, 6, 2]
mean_deviation = (7 + 1 + 6 + 2) / 4 # Equals 4
2 Likes
I think it should be pointed out that the sum of differences from the mean will always equal to zero. The lesson and exercise makes it seem like it’s almost by luck we have negative differences that we need to get rid of when in reality you will always end up with negative and positive differences that sum to zero due to the nature of arithmetic mean.
Mathematically it can be proven as:
The sum of the differences from the mean for each of the n data point Xi is
Since the mean is calculated as sum of the data points divided number of data points n, it can be simplified as followed:
