Question

A process that fills packages is stopped whenever a package is detected whose weight falls outside the specification. Assume that each package has probability 0.02 of falling outside the specification and that the weights of the packages are independent. Find the mean number of packages that will be filled before the process is stopped. Numeric Response

Answer 1

Answer:

The mean number of packages that will be filled before the process is stopped is 50.

Step-by-step explanation:

For each package, there are only two possible outcomes. Either it fails outside the specifications, or it does not. Packages are independent. This means that we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Number of trials expected for n sucesses

Also called inverse binomial distribution, is given by:

$E = \frac{n}{p}$

In which p is the probability of a success in a trial.

Assume that each package has probability 0.02 of falling outside the specification and that the weights of the packages are independent.

This means that $p = 0.02$

Find the mean number of packages that will be filled before the process is stopped.

This is when $n = 1$, as the process is stopped when a package is outside the specifications. So

$E = \frac{n}{p} = \frac{1}{0.02} = 50$

The mean number of packages that will be filled before the process is stopped is 50.