Average- how to quickly calculate

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Average- how to quickly calculate

by rishijhawar » Tue Jul 19, 2011 7:27 am
Just thought to check if there is any efficient way to calculate average of a set of non-consecutive numbers (say 993, 994, 996, 997, 998, 1001, 1001, 1002, 1004 and 1004).
[spoiler]Ans: 999[/spoiler]

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by hagan » Tue Jul 19, 2011 3:55 pm
my eye is on this post, am waiting for response from our experts
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by goalevan » Tue Jul 19, 2011 5:35 pm
Yes, there is actually a very easy way. When you take the mean of a set of numbers, the "residuals" will add to zero.

For example: the average of 7, 8, 9, 10, 11 is 9, and:

7 - 9 = -2
8 - 9 = -1
9 - 9 = 0
10 - 9 = 1
11 - 9 = 2

(-2) + (-1) + 0 + 1 + 2 = 0

This property allows you to start with a "base" number and calculate the residuals from that number to determine where the mean is.

From your example, we can find the average by taking a base number, say 1000, and calculating the sum of the residuals:

993 - 1000 = -7
994 - 1000 = -6
996 - 1000 = -4
997 - 1000 = -3
998 - 1000 = -2
1001 - 1000 = 1
1001 - 1000 = 1
1002 - 1000 = 2
1004 - 1000 = 4
1004 - 1000 = 4

The sum of these numbers is (-7) + (-6) + (-4) + (-3) + (-2) + 1 + 1 + 2 + 4 + 4 = -10

1000 is thus not the mean, but you know that the mean -10/10 lower than 1,000 (since there are 10 numbers that make up the average). The average is 1000 - 1 = 999. This is MUCH easier than trying to sum the numbers and divide.

Another quick example. The average of 650, 662, 668, and 676. Let's take 660 as a base:

650 - 660 = -10
662 - 660 = 2
668 - 660 = 8
676 - 660 = 16

Sum of residuals: 16
Number of observations: 4
+16/4 = 4

The average is thus 660 + 4 = 664.

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by blaster » Tue Jul 19, 2011 8:55 pm