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Suppose a truck has a capacity of 100 and order sizes to be filled are normal distributed with mean 95 and standard deviation of 10. There is about 30% chance that capacity is exceeded. In this case the truck is sent off with a load of 100. How do you calculate the average truck load?

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You can read about truncated normal distributions on Wikipedia.

Using $f(x)$ for the normal probability density with $\mu=95$ and $\sigma=10$, and $F(x)$ for the cumulative distribution, what you are looking for is

$$\int_{-\infty}^{100} x f(x) dx + 100\int_{100}^{\infty} f(x) dx$$

and the first integral comes to $ \mu F(100) - \sigma^2f(100) $ while the second integral comes to $100(1-F(100)) $, giving a sum of $$100 - (100-\mu)F(100) - \sigma^2 f(100)$$ $$ \approx 100 - 5 \times 0.6914625 - 100 \times 0.03520653 $$ $$\approx 93.02203. $$

Strictly you should also truncate if the normal distribution gives a load of less than zero, but in this case that part of the calculation would be lost in the rounding.

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