25th percentile of the sum of 500 samples coming from a given probability deistribution I came across this problem that was modified from a business case study:

An environmental start-up recycles catalytic converters. It estimates that the worth of one catalytic converter has the following probability distribution: 1200\$ with probability 0.35, 1800\$ with probability 0.50, and 2400\$ with probability 0.15. The start-up is offered a load of 500 catalytic converters, and the start-up needs to assess the worth of the entire load to decide whether to purchase it (assuming that the worth of any one converter is independent of the worth of the others). To ensure a profit, the start-up would not pay a price higher than the 25th percentile of the probability distribution associated with the worth of the load. (In other words, they would agree to a price only if there is a $75 \%$ chance that the load is worth more than that price.) Following this rule, what is the maximum price that the start-up would be willing to pay for the load of 500 catalytic converters?

If we define $X$ to be the random variable for the worth of one converter and $Y$ to be the random variable for the worth of 500 converter, then $E[Y]=E[500X]=500E[X]=500(1200(0.35)+1800(0.5)+2400(0.15))=840000$ and $\sigma(Y)=\sigma(500X)=500\sigma(X)=203469.89$.
My first instinct to answer the question was use the observation that the 25th percentile for $X$ has value 1200 and multiply this to 500. However, this is probably wrong.
My second instinct is that $$\text{25th percentile of the total worth} =\text{expected value} + (z-\text{value of 25th percentile})(\text{standard deviation}).$$
However, I still think that this is wrong. Since we have a sample of 500 here, I think the central limit theorem must come into play somehow. How does one compute for the 25th percentile of the total worth of 500 converters?
 A: I will use thousands of dollars throughout, in order to keep numbers from
getting too large and to avoid writing more digits than precision supports.
Let $X$ be the value of a randomly chosen catalytic
converter. Then  $\mu_X = E(X) = 1.68$ thousand dollars and
$\sigma_X = SD(X) = 0.407$ thousand dollars. [Computations in R.]
kd = c(1.2, 1.8, 2.4)
pr = c(.35, .50, .15)
mu.x = sum(kd*pr); mu.x
[1] 1.68

va.x = sum(pr*(kd-mu.x)^2); va.x
[1] 0.1656
sg.x = sqrt(va.x);  sg.x
[1] 0.4069398

Then the total value $T$ of 500 converters should
have $\mu_T=E(T) = 500\mu_X = 840$ thousand dollars and
$\sigma_T = SD(T) = \sqrt{500\sigma_X^2} =  9.0995$
thousand dollars.
mu.t = 500*mu.x;  mu.t
[1] 840
sg.t = sqrt(500*va.x);  sg.t
[1] 9.099451

The random variable $X$ is highly discrete, but it
is probably OK to assume that $T$ is approximately
$\mathsf{Norm}(\mu_T = 840, 
\sigma_T = 9.09945)$ in thousands of dollars.
Then the lower quartile of $T$ would be the lower
quartile of this normal distribution: 833,863 thousand dollars.
qnorm(.25, 840, 9.09945) 
[1] 833.8625

Simulation.
We can simulate 100,000 batches of 500 converters
to see how well results match with those above. With
100,000 iterations we can expect at at least two significant digits of accuracy, so results
match within the margin of simulation error.
In particular the normality assumption does seem OK,
set.seed(125)
t = replicate(10^5, sum(sample(kd, 500, rep=T, p=pr)))
mean(t);  sd(t);  quantile(t, .25)
[1] 840.0297
[1] 9.084199
25% 
834 


 hist(t, prob=T, col="skyblue2", 
      xlab="Thousand Dollars", main="Simulated Batches")
  curve(dnorm(x, mean(t), sd(t)), add=T)
  abline(v = 834, col="red", lwd=2)

Note: The simulated lower quartile 834 thousand dollars, might be
more accurate than the lower quartile 833.8625 thousand dollars of the theoretical
normal distribution. Even though the histogram does not show that the distribution of $T$ is discrete; among the 100,000
simulated values of $T$ there were only $124$ uniquely different values.
 length(unique(t))
 [1] 124

If this is a textbook exercise in means, variances, and
quartiles, then \$833,862.50 is likely the expected
answer. But if it's a real-life problem, in which the lower
quartile is set as protection that someone may have skimmed
many of the more valuable converters from some batches,
then it may be better to use the simulated value \$ 834,000.
