Each of the batteries in a collection of 40 batteries is equally likely to be either a type $A$ or a type $B$ battery. Type $A$ batteries last for an amount of time that has mean $50$ and standard deviation $15$; type $B$ batteries last for an amount of time that has mean $30$ and standard deviation $6$.
Question - Approximate the probability that the total life of all $40$ batteries exceeds $1700$.
Edited answer
Let Y be the type of battery, where $Y = 1$ for Type $A$ battery and $Y =2$ for Type $B$ battery.
$E[X] = E[E[X|Y]] = E[X|Y=1]P[Y=1] + E[X|Y=2]P[Y=2] = (50*0.5) + (30*0.5) = 40$
$Var[X] =E[X^2] - [E[X]]^2$
Hence, we need to find $E[X^2]$, $E[X^2] = E[E[X^2|Y]] = E[X^2|Y=1]P[Y=1] + E[X^2|Y=2]P[Y=2] = (15^2+50^2)(0.5)+(30^2+6^2)(0.5) = 1830.5$
Therefore ,$Var(X) = 1830.5 - 40^2 = 230.5 $
Knowing the variance and mean, How do I do the next part?
The key question to ask is, if you knew in advance how many batteries were of each type, how would you model the resulting distribution of the total lifetime?
In other words, suppose you were able to tell that $Y$ batteries were of type $A$, and $40 - Y$ batteries were of type B. Then you could number the batteries so that $$T_1, \ldots, T_Y \sim \operatorname{Normal}(\mu_A = 50, \sigma_A = 15)$$ represent the random lifetimes of the Type $A$ batteries, and $$T_{Y+1}, \ldots, T_{40} \sim \operatorname{Normal}(\mu_B = 30, \sigma_B = 6)$$ represent the random lifetimes of the Type $B$ batteries. Now, because the sum of independent (but not necessarily identically distributed) normal distributions is itself normal, we can model the total random lifetime conditioned on $Y$: $$S \mid Y \sim \operatorname{Normal}(\mu , \sigma)$$ where $$\mu = \mu_A Y + \mu_B (40-Y) = 50Y + 30(40-Y) = 20Y + 1200,$$ and $$\sigma^2 = \sigma_A^2 Y + \sigma_B^2 (40-Y) = 225 Y + 36(40-Y) = 189Y + 1440.$$ Note that the variance is additive, not the standard deviations themselves. So the conditional probability that the total lifetime of all $40$ batteries is more than $1700$, given that $Y$ batteries are of type $A$, is $$\Pr[S > 1700 \mid Y] = \ldots.$$ Then the unconditional (marginal) probability is computed as $$\Pr[S > 1700] = \sum_{y=0}^{40} \Pr[S > 1700 \mid Y = y] \Pr[Y = y],$$ where $$Y \sim \operatorname{Binomial}(n = 40, p = 1/2).$$ This computation will give an exact result, not an approximation. I leave it to you to continue this line of reasoning, and apply the appropriate approximation(s) if needed, to obtain a result.