Lets say, we're calculating averages of test scores:
Starting Test Scores: 75, 80, 92, 64, 83, 99, 79
Average = 572 / 7 = 81.714...
Now given 81.714, is there a way to add a new set of test scores to "extend" this average if you don't know the initial test scores?
New Test Scores: 66, 89, 71
Average = 226 / 3 = 75.333...
Normal Average would be: 798 / 10 = 79.8
I've tried:
Avg = (OldAvg + sumOfNewScores) / (numOfNewScores + 1)
(81.714 + 226) / (3 + 1) = 76.9285
Avg = (OldAvg + NewAvg) / 2
(81.714 + 79.8) / 2 = 80.77
And neither come up the exact average that it "should" be. Is it mathematically possible to do this considering you don't know the initial values?
3 Answers. Show activity on this post. i.e. to calculate the new average after then nth number, you multiply the old average by n−1, add the new number, and divide the total by n. In your example, you have the old average of 2.5 and the third number is 10.
There are three main types of average: mean, median and mode. Each of these techniques works slightly differently and often results in slightly different typical values. The mean is the most commonly used average. To get the mean value, you add up all the values and divide this total by the number of values.
How to Calculate Average. The average of a set of numbers is simply the sum of the numbers divided by the total number of values in the set. For example, suppose we want the average of 24 , 55 , 17 , 87 and 100 . Simply find the sum of the numbers: 24 + 55 + 17 + 87 + 100 = 283 and divide by 5 to get 56.6 .
To find the arithmetic mean of a data set, all you need to do is add up all the numbers in the data set and then divide the sum by the total number of values.
You have to know the number of test scores in the original set and the old average:
newAve = ((oldAve*oldNumPoints) + x)/(oldNumPoints+1)
Say, you have 2 blocks of scores:
1st: n scores with average = a1
2nd: m scores with average = a2
then average of total scores equals:
a1*(1.0*n/(m+n))+a2*(1.0*m/(m+n))
In case you want just add 1 score (a2) to existing set formula becomes
a1*(n/(n+1))+ a2/(n+1)
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