Coal is carried from a mine in West Virginia to a power plant in New York in hopper cars on a long train. The automatic hopper car loader is set to put 75 tons of coal into each car. The actual weight of coal loaded into each car is normally distributed, with mean of 75 tons and standard deviation of 0.8 ton.
I have $\mathcal{N}(75, 0.8)$. I would like to find, Among 20 randomly chosen cars, the probability that more than 2 cars will be loaded with less than 74.4 tons of coal.
My approach as follow:
First I have found $P(X<74.4) = 0.2266$
Then I treat it as $Bin(20, 0.2266)$
Hence, let Y be number of car loaded with less then 74.4 tons of coal.
I need $P(Y>2) = 1 - P(Y=0) - P(Y=1)$
I would like to know that, is my approach correct? i.e. treat it as binomial distribution.