Stats help required
probability
Not probability
"infect"?, who wrote this question
infact*
@modphysnoob
GM is geometric mean?
(wrong_GM*10)-wrong_number+right number --------------------------------------- 10
thanks
@modphysnoob Please can you tell me where you got this formula from?
@kropot72 Please help
If the Wrong GM is 12.9,What is the wrong number here?
When any number of quantities are in a geometric progression, the terms intermediate between the first and the last are called the geometric means between those two terms. It is incorrect to state a single value for the geometric mean of 10 observations. There are 8 values of geometric mean.
You mean to say the question is wrong?
@kropot72
Okay so the geometric mean is going to be: \[ \Large \sqrt[n]{\Pi_i^n x_i} \]Right?
So first we take the GM to the power of 10, to get it in the form: \[\Large \Pi_i^nx_i \]
Then we divide out our wrong value and multiply in the right value.
Then we take it to the power of \(1/10\) (the 10th root) to get it back as a geometric mean.
@hba It's all about reverse engineering.
@wio I am not actually get it :(
Okay do you know what geometric mean is?
Yes GM=\[\sqrt[n]{(x_1+x_2+........x_n)}\]
Wait are you sure about that?
Yeah
\[\Large GM = \left(\prod_{i=1}^n a_n \right)^{(1/n)} \]
It's not a sum, you are multiplying them
Oh yeah sorry
Okay so \[ \Large GM^n =\prod_{i=1}^na_i \]
If we want to say that \(a_k\) is wrong and should be replaced with \(b\), then: \[\Large \frac{b}{a_k}GM^n =\frac{b}{a_k}\prod_{i=1}^na_i \]Just divide it out and multiply in the new one.
I am sorta getting it
\[ \Large GM_2 = \left(\frac{b}{a_k}GM^n \right)^{1/n} \]
What is this GM_2 for?
The new geometric mean after replacing one of the values.
I am really sorry i may sound stupid because Stats is something new for me.
You don't get what I did?
Well lemme tell you what i got.Wait a sec
i did this, |dw:1362302169899:dw|
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