[SOLVED] Calculate peak of Gaussian distribution from known integral

Jun 2010
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For example, there is a website that has 100,000 pageviews every day.

We assume the distribution of these pageviews fits a Gaussian distribution.

How would the theoretical peak pageviews per second be calculated?
 

CaptainBlack

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Nov 2005
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erehwon
For example, there is a website that has 100,000 pageviews every day.

We assume the distribution of these pageviews fits a Gaussian distribution.

How would the theoretical peak pageviews per second be calculated?
Please consider rewording this problem to make it unambiguous what has a Gaussian distribution, what is its meand, ...


CB
 
Jun 2010
5
0
For each day, the number of pageviews per second is assumed to fit a Gaussian Distribution, with the peak occurring at 12pm and evenly distributed either side of this.

It is required to determine the peak number of pageviews per second given this distribution of pageviews.
 

CaptainBlack

MHF Hall of Fame
Nov 2005
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erehwon
For each day, the number of pageviews per second is assumed to fit a Gaussian Distribution, with the peak occurring at 12pm and evenly distributed either side of this.

It is required to determine the peak number of pageviews per second given this distribution of pageviews.
I think as asked here this still does not have an answer.

One possible interpretation is:

If the mean number of pageviews per second varies through the day with a bell shaped graph which is of approximatly Gaussian form, with peak at 12:00 (that is mid-day) what is the peak mean pageviews?

(note there is no mention of a distribution here since there is no probability distribution involved)

Even this interpretation has no answer as a relevant peice of data is missing, which is some sort of specification of the spread of the Gaussian.

CB
 

CaptainBlack

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Nov 2005
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erehwon
Do you mean the variance?
It does not have a variance as it is not a probability distribution, I mean the width of the peak measured in some appropriate manner, which may be in the for of the square root of the second moment of the area under the curve, which is the analogue of the standard deviation. But the full width at half maximum would do.

CB
 
Jun 2010
5
0
Argh, of course. Sorry - it's Friday afternoon - I had it clear in my mind, I just didn't write it down.

Ideally I'd like to derive a function to calculate the peak value in terms of σ so that it would be possible to derives peak values for various durations of the peak.
 

CaptainBlack

MHF Hall of Fame
Nov 2005
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erehwon
Argh, of course. Sorry - it's Friday afternoon - I had it clear in my mind, I just didn't write it down.

Ideally I'd like to derive a function to calculate the peak value in terms of σ so that it would be possible to derives peak values for various durations of the peak.
\(\displaystyle f(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{-(x-\overline{x})^2/(2\sigma^2)}\)

the maximum occurs at \(\displaystyle x=\overline{x}\) (\(\displaystyle \sigma\) in units of minutes), so is:

\(\displaystyle \max(f(x))=\frac{1}{\sqrt{2\pi}\ \sigma}\)

which is the case when the total number is 1, so we multiply this by 100,000 to get the peak views per minute.

CB
 
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Jun 2010
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Great - thanks! But surely σ should be in units of time as it determines the width of the bell?