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I have a Lognormal distribution generated in the following way.

List of parameters set by me (the user):

  • mean = 180
  • median = 120
  • truncation point = 1800

From this parameters we generate a NORMAL distribution:

$Normal(\mu,\sigma)$ where $\mu = ln(median)$ and $\sigma = \sqrt{2 \times ln (\frac{mean}{median})}$

Then, to generate the LOGNORMAL distribution we take the exponent value of the Normal distribution.

Now, i have the truncation point: I want to know if in my LogNormal distribution there is more than the 5% of data that are bigger then the truncation point. Is it possible?

Thank you

1 Answers 1

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If I understand you correctly, you are asking whether more than 5% of the values of the given lognormal distribution are greater than 1800. This is the same as asking whether a $N(\mu,\sigma)$ random variable is greater than $\ln 1800$ (take the log of both sides). This is the same as asking whether a $N(0,1)$ value is greater than $\frac{\ln 1800 - \mu}{\sigma} \approx 2.55$ (check my fat fingered calculator exercise). As only about 1.5% of the N(0,1) distribution is greater than 2.55 (again check my computations) the answer is no, there is less than 5% of the distribution greater than 1800.

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    I'd certai$n$ly cop to my calculatio$n$s bei$n$g wro$n$g. Other than the arithmetic, it seems correct.2011-07-09