Question

log10X is normally distributed with mean 7 and variance 3, log10Y is also normally distributed with mean 3 and unit variance. If distribution of X and Y are independent, find the probability of 1.202<X/Y<83180000. (Given that log10 1202=3.08, log10 8318=2.92 )

Answer 1

when X and Y are independent ,

$P(X/Y)=P(X)\\$

Therefore,

$P(1.202<X/Y<83180000)=P(1.202<X<83180000)=P\\$

consider the log normal distribution,

$P=P(log_{10}{1.202}<log_{10}{X}<log_{10}{83180000})\\$

$log_{10}{1.202}=log_{10}{(\frac{1202}{1000})}=log_{10}{1202}-3=0.08\\ log_{10}{83180000}=log_{10}{(8318*10000)}=log_{10}{8318}+4=6.92\\$

Then,

$P=P(0.8<log_{10}{X}<6.92)\\ normalize\ the\ value,\\ z=\frac{X-\bar{X}}{\sigma}\\ z_1=\frac{0.8-7}{\sqrt{3}}=-3.5796\\ z_2=\frac{6.92-7}{\sqrt{3}}=-0.0462\\ P=P(-3.5796<z<-0.0462)\\ P=P(-3.5796<z<0)-P(-0.0462<z<0)\\ P=P(0<z<3.5796)-P(0<z<0.0462)\\ using\ calculator\ or\ table,\\ P=0.4998-0.0184=\bold{0.4814}\\$