0
$\begingroup$

A set of numbers is transformed by taking the log base 10 of each number. The mean of the transformed data is 1.65. What is the geometric mean of the untransformed data?

 Approach used so far:
    i) So basically a * b * c ...*tn = 10^(1.65*n) where a,b,c and so on are the values in the data set taken.
    ii) I have to find (a * b * c ... *tn^1/2 = 10^(1.65/2*n)
$\endgroup$
1
$\begingroup$

Lets define the N numbers in your set with $x_i$, where $i \in \{1,...,N \}$.

Since $\sum_{i=1}^{N} \log_{10}x_i$ = $\log_{10}(\prod_{i=1}^{N} x_i)$ and the mean of the transformed data is 1.65, it follows that \begin{equation} \frac{\log_{10}(\prod_{i=1}^{N} x_i)}{N} = 1.65 \Leftrightarrow \prod_{i=1}^{N} x_i = 10^{N \cdot 1.65} \end{equation} Taking by the power of $\frac{1}{N}$ on both sides yields you the geometric mean of the untransformed data: \begin{equation} \left(\prod_{i=1}^{N} x_i\right)^{1/N} = 10^{1.65} \end{equation}

$\endgroup$
0
$\begingroup$

I would add this as a comment if I had 50 rep.

Remember your basic log rules. You have the sum of logs, can you think of how that can be expressed differently? Also if this is self study or homework please use the self study tag.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.