In an example, I am told that
$$\int_0^T{e^{-bt}}\,dt=\frac{1}{b}(1-e^{-bT})$$
is $n(0,T)$, i.e., which I took to mean normally distributed with mean of $0$ and variance of $T$.
Note: The value of $b$ and $T$ are unspecified in the problem.
I know the PDF and CDF of the normal and standard normal distributions and I didn't make the pattern recognition. Can you help me see how it is a normal distribution with mean of $0$ and variance of $T$?
Thanks in advance.
Update: Since I am in agreement with Chinny84 and Did that this function does not qualify as being normally distributed, I have chosen to interpret the author's notation as not meaning a normal distribution, but as a simplifying substitution rather than implying a normal distribution - despite the unfortunate use of notation that normally (pun a natural consequence) denotes the normal distribution. And that approach does work (I can't elaborate). The source is a tiny part of a larger problem statement on an exam. I'll award Chinny84 the answer for the help his answer will provide others who come after. Thanks again for your interest/help.
The above expression is not a normal which would have a form $$ \frac{1}{\sqrt{2\pi b}}\mathrm{e}^{-0.5\left(\frac{x-a}{b}\right)^2} $$
Where a is the mean and b is the standard dev.
The above is not even an exponential distribution since that would have the form $\frac{b}{1-\mathrm{e}^{-bT}}\mathrm{e}^{-bt}$ for [0,T).
To find mean and variance you would compute the corresponding moments ie integrals for a pdf. However, if you can see that it is clearly of Gaussian form then by inspection you can determine the mean and variance.
I always give warning when reading a response (like my one) and double check for yourself.
Hope it helps.
Rob