Question

4. A soft-drink machine is regulated so that it discharges an average of 200 mls per cup. If the amount of drink is normally distributed with a standard deviation equal to 15 mls. Below what value do we get the smallest 30% of the cups? 

Answer 1

Let X represent the amount of drink distributed.

$\mu = 200 \\ \sigma=15$

The value if we get the smallest 30% of the drinks

$P(X<x) = 0.3\\ P(Z< \frac{x-200}{15}) = 0.3 \\ P(Z< -0.68) = 0.3\\ \frac{x-200}{15} = -0.68 \\ x = -0.68 \times 15 + 200 \\ x=189.95$