Question

A quality measurement for cabinet manufacturers is whether a drawer slides open and shut easily. Historically, 2% of drawers fail the easy slide test. A manufacturer samples 10 drawers from a batch. Assuming the chance of failure is independent between drawers, what type of distribution could be used to model the number of failed drawers from the sample of 10?

Answer 1

Binomial distribution could be used to model the number of failed drawers from the sample of 10.

Explanation:

We are given the following information:

We treat drawers fail the easy slide test as a success.

P(drawers fail the easy slide test) = 2% = 0.02

The chance of failure is independent between drawers.

A manufacturer samples 10 drawers from a batch.

Since,

• The experiment consists of 10 repeated trials.

• Each trial can result in just two possible outcomes - drawers fail the easy slide test or drawer passes the easy slide test.

• The chance of failure is independent between drawers.

Then the number of drawers that fail the easy slide test follows a binomial distribution, where

`P(X=x) = \binom{n}{x}.p^x.(1-p)^(n-x)`

where n is the total number of trails, x is the number of success, p is the probability of success.

Here, the parameters of binomial distribution are

n = 10, x = number of failed drawers(discrete values, x = 0, 1,...,10), p = 0.02

`P(X=x) = \binom{10}{x}(0.02)^x.(0.98)^(10-x)`