Question

If log10X is normally distributed with mean 4 and variance 4, find the probability of 1.202<X<83180000.(Given that log10 1202=3.08, log10 8310=3.92 )

Answer 1

Let require probability, $P=P( 1.202<X<83180000)$

$P( 1.202<X<83180000)$ is convert to $\log_{10}X$.

Then ,

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

$\log_{10}{1.202}=\log_{10}{\frac{1202}{1000}}=\log_{10}{1202}-\log_{10}{1000}\\$

$=3.08-3=0.08$

$\log_{10}{83180000}=\log_{10}{8318*10^4}$

$=\log_{10}{8318}+\log_{10}{10000}\\$

$=3.92+4$

$=7.92$

$P=P(0.08<\log_{10}X<7.92)$

Then normalize the values,

$z=\frac{x-mean}{standard\ deviation}\\ z_1=\frac{0.08-4}{2}=-1.96\\ z_2=\frac{7.92-4}{2}=1.96\\$

$P=P(-1.96<z<1.96)\\ P=P(-1.96<z<0)+P(0<z<1.96)\\ P=2*P(0<z<1.96)\\$

using calculator or table,

$P=2*0.475\\ \bold{P=0.95}$