How to solve this sum of normal distribution question? Assume the weight of a person follows a normal distribution N(71,7). What is the probability of 4 people weighing more than 300kg?
I tried solving this by multiplying the values by 4, so it'd be N(284,28). I converted that into $x=284+28z$ which lead to  $z=\frac{x-284}{24}$.
To solve for $P(x>300)$, I converted it around to $ 1-P(x<300)$ $$ 1-P\left(\frac{300-284}{24}\right) $$
$$ 1-P\left(z<\frac{2}{3}\right) $$
Then i used the normal distribution table and got the following result: $1-0.2546 = 0.745$, however I have a suspicion I started from the wrong track and this result isn't correct.
This was an exam question at a college statistics class
 A: Your mistake is that variances add, not standard deviations. If $7$ is the one-person variance, work with $300=284+\sqrt{28}z$; if $7$ ($49$) is the one-person srandard deviation (variance), work with $300=284+\sqrt{4\times49}z=284+14z$.
A: I assume heights are independently normally distributed with mean $\mu = 71$ and standard deviation $\sigma = 7.$ [Sometimes the second parameter is the variance and sometime the variance, you should always say which convention you are using.]
Then $S = X_1+X_2+X_3+X_4$ has $E(S) = E(X_1)+\cdots+E(X_4) = 4(71) = 284.$
and (because of independence) $V(S) = V(X_1)+\cdots+V(X_4) = 4(7^2) = 196.$
So $P(S > 300) = 1 - P(S_4 \le 300 = 0.1265.$ Computation in R (where the second parameter is the SD). You can get approximately the same answer by standardizing and using printed tables of the standard normal CDF (which
involves some rounding.).
1 - pnorm(300, 284, sqrt(196))
[1] 0.126549

So an elevator with safe weight load 300, serving a population with
weights distributed $\mathsf{Norm}(\mu = 71, \sigma=7)$ should have
a limit of three (not four) passengers.
Simulation: With a million sums of four we get $0.1266\pm 0.0007.$
set.seed(1120)
x1 = rnorm(10^6, 71, 7)
x2 = rnorm(10^6, 71, 7)
x3 = rnorm(10^6, 71, 7)
x4 = rnorm(10^6, 71, 7)
s = x1 + x2 + x3 + x4

summary(s);  sd(s)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  217.9   274.6   284.0   284.0   293.4   351.7 
[1] 13.99282
mean(s > 300)
[1] 0.126589         # aprx P(S > 300) = 0.126549
2*sd(s > 300)/1000
[1] 0.0006650243     # aprs 95% margin of simulation err    


hist(s, prob=T, br=30, col="skyblue2", main="Sim Dist'n Sum of Four")
 curve(dnorm(x, 284, 14), add=T, col="red", lwd=2)
 abline(v=300, lty="dotted", lwd=2)

You can get a good approximation by standardizing and using printed tables of
the standard normal CDF (a proxess which requires some rounding).
