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

$P=P(1.202<X<83180000)$

let log normal distribution,

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

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

convert to standard form using by

$z=\frac{x-\bar{x}}{\sigma}$

$P=P(\frac{0.08-4}{2}<Z<\frac{7.92-4}{2})\\ P=P(\frac{0.08-4}{2}<Z<\frac{7.92-4}{2})\\ P=P(-1.96<Z<1.96)\\ P=P(-1.96<Z<0)+P(0<Z<1.96)\\$

using calculator or table,

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