Question

If is normally distributed with mean 4 and variance 4, find the probability of 1.202<X<83180000. (Given that log10 1202 =3.08,log10 8318=3.92)

Answer 1

$P=P(1.202<X<83180000)$ is similar in log normal as,

$P=P(\log_{10}1.202<\log_{10}X<\log_{10}83180000)\\ P=P(\log_{10}\frac{1202}{1000}<\log_{10}X<\log_{10}(8318*10^4))\\ P=P(3.08-3<\log_{10}X<3.92+4)\\ P=P(0.08<\log_{10}X<7.92)\\$

converting to normal form using,

$z=\frac{x-mean}{standard\ deviation}=\frac{x-4}{2}$

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

using calculator or table,

$P=0.475+0.475\\ \bold{P=0.95}$