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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

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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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  • $\begingroup$ Thank you! Now i try to verify if this procedure satisfies my problem. $\endgroup$
    – Maurizio
    Commented Jul 9, 2011 at 9:57
  • $\begingroup$ Maybe your calc is wrong: $\frac{ln 1800 - \mu}{\sigma} = 3.00722 \approx 3.00$ than with Z table i check that P(z>3.00) = [total area] - [area to the left of 3.00] = 1 - 0.99865 = 0.00135 that is less than 5%. Is it right? $\endgroup$
    – Maurizio
    Commented Jul 9, 2011 at 10:10
  • $\begingroup$ I'd certainly cop to my calculations being wrong. Other than the arithmetic, it seems correct. $\endgroup$
    – deinst
    Commented Jul 9, 2011 at 14:23

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