Finding probability using central limit theorem

The weight (in kg) in a population has a mean of 70 & a standard deviation of 10kg. Suppose that a passenger vechicle that can seat 40 such people can safely carry 3000kg. According to the central limit theorem, when carrying 40 passengers what is the probability that the total weight exceeds 3000 kg?

So I did..... 3000-2800/400 = P>0.5. This isn't the right answer though. Can anyone show me what I've done wrong.
Thankyou so much in advance!

You should express your answer in terms of Xbar (the sample mean) rather than the total weight.
Also, your standard deviation needs to be divided by sqrt(n), since we are taking a sample.

Let Xk be the weight of the kth person, where k is from 1 to 40.
Then Xk ~ Normal(70, 10^2).
So by the central limit theorem, since n = 40 is large, we have Xbar ~ Normal(70, 10^2/40).
That is, the mean of Xbar is 70kg and the standard deviation of Xbar is 10/sqrt(40) kg.

We want:
P(X1 + X2 + ... + X40 > 3000)
= P(Xbar > 3000/40)
= P(Xbar > 75)
= P(Z > (75 - 70)/(10/sqrt(40))
= P(Z > 3.1623)
= 0.0008

So it's very unlikely (0.08% chance) that the total weight would exceed 3000kg.

With n=40, Xbar~N(70, 10^2/40)
xbar=3000/40=75.
P{X>75}=P{Z>z}, where z=(75-70)/(10/√40)=3.162
P{Z>3.162}=P{Z<-3.162}=0.0008