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Mean of the Means

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by MBA.Aspirant » Sat Jul 02, 2011 2:48 pm
When do you take the average of the averages given, and when do you do their sum then divide by the no. of terms?

Simple example:

The average of x and y is 10. The average of y and z is 18. And the average of x and z is 20. What is the average of x, y and z?"

you can easily add the sums, divide by 2 then by 3. But the OA takes the average of these averages, which is the same in calculation but I don't get the concept underlying it.

Thanks
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by goalevan » Sat Jul 02, 2011 4:38 pm
It's just a slightly different way of making the calculation. The way you calculated, you just multiplied by 2 to get the sum for each of the three statements before you added them. The OA adds them before it multiplies out the 2.

Adding the statements:
x + y = 20
y + z = 36
x + z = 40
--------------------
2(x + y + z) = 96
(x + y + z) = 48
(x + y + z)/3 = 16

Taking the average of the averages, you leave the /2 in each term:
[(x + y)/2 + (y + z)/2 + (x + z)/2] / 3 = (20/2 + 36/2 + 40/2) / 3
[2(x + y + z) / 2] / 3 = 48/3
(x + y + z)/3 = 16

Both methods just happen to bring you to the answer.
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by MBA.Aspirant » Sat Jul 02, 2011 5:41 pm
Thanks man. I know by calculation that both are same. I want to know the concept behind taking the avg of the averages here.
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by MBA.Aspirant » Sun Jul 03, 2011 1:19 am
bump
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by Frankenstein » Sun Jul 03, 2011 6:06 am
Hi,
There is no underlying concept in this. It is just the same thing interpreted in two ways. So, no need to worry about any concept here.
Cheers!

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by MBA.Aspirant » Sun Jul 03, 2011 10:49 am
I just learned from Kaplan that you can take the average of the averages when the sets have equal no. of terms. Like in two sets of 60 items combined. Avg of new Set = Avg Set A +Avg Set B/ 2

If the no. of terms aren't equal in both sets, then it's weighted average.

now can someone tell me how this apply to this example?

The average of x and y is 10. The average of y and z is 18. And the average of x and z is 20. What is the average of x, y and z?
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by amit2k9 » Sun Jul 03, 2011 10:59 pm
2(x+y+z) = 20+36+40 = 96
x+y+z = 48

thus (x+y+z)/3 = 48/3=16
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