Question

A coffee machine fills a cup automatically with a volume of coffee normally distributed with expectation 2.2dl and standard deviation 0.3dl. The cups used can take a volume of coffee with expectation value 2.5 dl and standard deviation 0.45 dl. The volume of the coffee machine drops and the volume of a cup is independent.

What is the probability of it accidentally overflowing with a cup?

Answer 1

Let coffee filled be represented by and volume of cup be

We are considering the probability of accidentally overflowing. Overflowing means that the coffee drops are more than the volume. That means therefore

So we need a distrbution of then we find the probability of .

This is a rule where E( X - Y) = E(X) - E(Y)

V( X - Y) =V(X) + V(Y) ..........there is a covariance term which is eliminate if X and Y are independent.

Therefore

P() = P(Z > )

= P( Z > 0.54)

= 1- P( Z < 0.54)

= 1 - 0.71045 ...........using normal distribution tables

P(overflowing) = 0.28955