If the weight $X$ of bags of cement is normally distributed with a mean of $40$ kg and a standard deviation of $2$ kg, how many bags can a delivery truck carry so that the probability of the total load exceeding $2000$ kg will be $5\%$?
The attempt at a solution:
Suppose $n$ bags are required. Their weights are iid normal variables, with mean $40$ kg and variance $4$ kg. Their sum is thus distributed $N(40n, 2 \sqrt{n})$. We can normalize this by defining $$Z = \frac{X - 40n}{2\sqrt{n}}$$ We have $X = 2000$ kg, and from a table of $\Phi(x)$ one obtains $Z = 1.645$, but the equation has no real solutions.
