Question

An automatic coffee machine is programmed so that it produced an average of 200 ml per cup. The amount of coffee follows a normal distribution with standard deviation equal to 10ml. if 25% of the drinks are below k value, find the value of k?

Answer 1

$P(Z<z)=0.25\to z=-0.6745.$

$\frac{k-\mu}{\sigma}=z.$

$\frac{k-200}{10}=-0.6745.$

$k=200-0.6745*10=193.255\;ml.$