The probability of the $\mathrm{SO_2}$ concentration over a forest is lognormally distributed with mean $1.9$ and standard deviation $0.9$. What is the probability that the concentration lies between $5$ and $10$?
I have tried to solve this the normal way by just standardizing the lognormal distribution: $$ \begin{eqnarray} P\left( 5 < Y < 10 \right) &=& P\left( \ln{5} <\ln{Y} < \ln{10} \right) \\ &=& P\left( \ln{Y} < \ln{10}) - P(\ln{Y} < \ln{5} \right) \\ &=& P\left( Z < \frac{\ln{10} - 1.9}{0.9}\right) - P\left(Z < \frac{\ln{5} - 1.9}{0.9} \right) \end{eqnarray} $$ with $Z$ being standard normally distributed. This equals approximately $$ \begin{eqnarray} P\left( Z < 0.44 \right) - P\left( Z < -0.32 \right) &=& P\left( Z < 0.44 \right) - P\left( Z > 0.32 \right) \\ &=& P\left( Z < 0.44 \right) - \left[1 - P\left( Z < 0.32 \right)\right] \;, \end{eqnarray} $$ and then using a standard normal table this should be equal to $$ P\left( 5 < Y < 10 \right) = 0.670 - (1 - 0.626) = 0.296 $$
But this is not the right answer. It should be $0.0087$ according to my book. So, where did I go wrong?