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Mathematics 7 Online
OpenStudy (hba):

Stats help required

OpenStudy (anonymous):

probability

OpenStudy (hba):

OpenStudy (hba):

Not probability

OpenStudy (anonymous):

"infect"?, who wrote this question

OpenStudy (hba):

infact*

OpenStudy (hba):

@modphysnoob

OpenStudy (anonymous):

GM is geometric mean?

OpenStudy (anonymous):

(wrong_GM*10)-wrong_number+right number --------------------------------------- 10

OpenStudy (hba):

thanks

OpenStudy (hba):

@modphysnoob Please can you tell me where you got this formula from?

OpenStudy (hba):

@kropot72 Please help

OpenStudy (hba):

If the Wrong GM is 12.9,What is the wrong number here?

OpenStudy (kropot72):

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.

OpenStudy (hba):

You mean to say the question is wrong?

OpenStudy (hba):

@kropot72

OpenStudy (anonymous):

Okay so the geometric mean is going to be: \[ \Large \sqrt[n]{\Pi_i^n x_i} \]Right?

OpenStudy (anonymous):

So first we take the GM to the power of 10, to get it in the form: \[\Large \Pi_i^nx_i \]

OpenStudy (anonymous):

Then we divide out our wrong value and multiply in the right value.

OpenStudy (anonymous):

Then we take it to the power of \(1/10\) (the 10th root) to get it back as a geometric mean.

OpenStudy (anonymous):

@hba It's all about reverse engineering.

OpenStudy (hba):

@wio I am not actually get it :(

OpenStudy (anonymous):

Okay do you know what geometric mean is?

OpenStudy (hba):

Yes GM=\[\sqrt[n]{(x_1+x_2+........x_n)}\]

OpenStudy (anonymous):

Wait are you sure about that?

OpenStudy (hba):

Yeah

OpenStudy (anonymous):

\[\Large GM = \left(\prod_{i=1}^n a_n \right)^{(1/n)} \]

OpenStudy (anonymous):

It's not a sum, you are multiplying them

OpenStudy (hba):

Oh yeah sorry

OpenStudy (anonymous):

Okay so \[ \Large GM^n =\prod_{i=1}^na_i \]

OpenStudy (anonymous):

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.

OpenStudy (hba):

I am sorta getting it

OpenStudy (anonymous):

\[ \Large GM_2 = \left(\frac{b}{a_k}GM^n \right)^{1/n} \]

OpenStudy (hba):

What is this GM_2 for?

OpenStudy (anonymous):

The new geometric mean after replacing one of the values.

OpenStudy (hba):

I am really sorry i may sound stupid because Stats is something new for me.

OpenStudy (anonymous):

You don't get what I did?

OpenStudy (hba):

Well lemme tell you what i got.Wait a sec

OpenStudy (hba):

i did this, |dw:1362302169899:dw|

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