Get the minimum value of passenger

Question

I have this statement:

The luggage weight of a commercial aircraft follows normal distribution $W(Weight) \sim N(20kg,4kg)$ .If the limit of The total luggage load of an aircraft carrying $100$ passengers is $2092.8kg$, So what is the probability that the limit is exceeded by these $100$ passengers?

My attempt was:

Assuming that each passenger will exceed the weight of their luggage by the same amount, then each passenger will exceed: $2092.8/100 = 20.928$

Thus, i need $P( W > 20.928) = k$

Normalize, $P(Z > 0.232) = 0.4083$

But the correct answer must be $0.01$, that is $1 - P(Z < 2.32)$, where $P(Z < 2.32) \approx 0.99$ And i don't know how to get this.

So, what is wrong in my attempt? Thanks in advance.

Answer 1

Use X as the sum of 100 iid normals, so X~N(2000kg,400kg). Then use X for the calculation. If the 4kg is the standard deviation, the Var(X) = 100(4^2) =1600 and X ~ N(2000 kg, 40 kg). This gets the correct answer.

Answer 2

What you need to find is the probability that the total weight of $100$ pieces of luggage, each of which independently has weight from a normal distribution with mean $20$ and standard deviation $4$, exceeds $2092.8$. Equivalently, you need to know the probability that the sample mean of a random sample of $100$ bags from the normal distribution $N(20,4)$ exceeds $209.28$. (It is not necessary for each of the $100$ bags to be above average in weight.) If you are at the point in your study where you’re expected to solve this problem, you should already be aware of the relationship between a distribution (the weights of bags) and the sampling distribution (the distribution of the means of samples of a given size) from that distribution.