# Lognormal distribution

mwendazimu
A labour force's annual incomes are lognormally distributed. If the labour force is arranged in order of decreasing annual incomes and the top 10% earns 37% of the total annual incomes, what proportion of the total annual income does the bottom 10% earn?

Kindly help on this one. It looks simple until you start solving and you realize that there is no mean or variance!
Also take note that all the information is provided. There is nothing missing in this question,

Pere Callahan
Welcome to PF

What is the density of the log normal distribution and how would you rephrase the given information in terms of this density?

Homework Helper
MHB
A labour force's annual incomes are lognormally distributed. If the labour force is arranged in order of decreasing annual incomes and the top 10% earns 37% of the total annual incomes, what proportion of the total annual income does the bottom 10% earn?

Kindly help on this one. It looks simple until you start solving and you realize that there is no mean or variance!
Also take note that all the information is provided. There is nothing missing in this question,

proportion bottom 10% = [TEX]1 - \Phi ( \Phi^{-1}(0.9) + \Phi^{-1}(0.37) - \Phi^{-1}(0.1) )[/TEX] = 1.28%

Last edited:
mwendazimu
proportion bottom 10% = [TEX]1 - \Phi ( \Phi^{-1}(0.9) + \Phi^{-1}(0.37) - \Phi^{-1}(0.1) )[/TEX] = 1.28%

I gave the exact solution and was given a big X. Could my professor be wrong?

Homework Helper
MHB
I gave the exact solution and was given a big X. Could my professor be wrong?

What I gave is my 2 cents, which I derived using the formulas given on wikipedia.
If you came out to the same answer that should be enough reason to go talk to your professor I guess.