| T O P I C R E V I E W |
| flmiguel |
Posted - 10/15/2014 : 09:37:47 AM
Origin Ver. and Service Release: OriginPro 8.6.0 (64-bit) b70 (Academic)
Operating System: Windows 7 Pro
Hello,
The image shows a histogram regarding the grain size distribution of some material and the respective lognormal fitting.
From this I want to express the representative value of my sample in form of "Mean value +- Error", where "Error" would be the standard deviation (at least for the Gaussian case).
I understand that, with the info of the fitting, this would be "xc +- w"
The mean value, or "xc" seems ok, but the standard deviation "w" is definitely wrong or at least my understanding of it. In this case, "w" is almost equal to the largest value used to build the histogram and it exceedes it clompetely when doing "xc + w".
Could anyone help me to understand this? Thank you very much.
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| 3 L A T E S T R E P L I E S (Newest First) |
| JacquelineHe |
Posted - 10/20/2014 : 05:33:35 AM
Hi,
If you want to calculate the standard deviation, please refer to this forum page:
http://www.originlab.com/forum/topic.asp?TOPIC_ID=19913
And our next new version will support calculate the standard deviation by distribution fit tool
Thanks
Jacqueline
OriginLab
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| flmiguel |
Posted - 10/16/2014 : 07:29:01 AM
quote:
Originally posted by JacquelineHe
In the Log-Normal function, the width value is calculated by log-normal "x" value.
As your sample, I assume the x is from 0.01 to 0.6.
width= ln(0.6)-ln(0.01) = ln(60) = 4.09434
So the w=0.6 is not the large value for this fitting result.
Thank you Jacqueline. I understand what you've said but that doesn't answer my question.
This w=0.6 corresponds to the natural logarithm of the grain diameter (my variable x of interest). What I want to know from the fitting, is the standard deviation of my grain diameter (not from its logarithm). If this were a gaussian fitting, I know that the sample standard deviation would be w/2 (=sigma). How can I obtain this value for the lognormal? Thanks! |
| JacquelineHe |
Posted - 10/16/2014 : 02:55:25 AM
Hi,
In the Log-Normal function, the width value is calculated by log-normal "x" value.
As your sample, I assume the x is from 0.01 to 0.6.
width= ln(0.6)-ln(0.01) = ln(60) = 4.09434
So the w=0.6 is not the large value for this fitting result.
Thanks
Jacqueline
OriginLab |
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