So by Property $X + Y \sim \mathcal{N}(1+1.5, 0.1^2 + 0.3^2) = \mathcal{N}(2.5,0.1)$
$$P(1<X+Y<1.3) = P\left(\frac {1.0-2.5}{(0.1)^.5}<Z<\frac {1.3-2.5}{(0.1)^.5}\right)$$
Which is equal to: P(-4.74< Z<-3.79) = 0.0001
Is this correct? I couldn't find a table for Z values as big as the one I calculated. I was just introduced to Normal distributions.
Temporarily added because of the disappearance of the orginal question:
We are told that $60\%$ of smurfs are female, and $40\%$ are male. Female smurfs have normally distributed height, mean $1$ m, standard deviation $0.1$. Male smurfs have normally distributed height, mean $1.5$ m, standard deviation $0.3$. If a smurf is selected at random, what is the probability that the smurf's height is between $1$ m and $1.3$ m?
Outline: Let $a$ be the probability that a randomly chosen female Smurf is between $1$ and $1.3$, and let $b$ be the corresponding probability for male Smurfs (Smurves?). Then our required probability is $(0.6)a+(0.4)b$.