A manufacturer of window frames knows from long experience that 5% of the production will have some type of minor defect that will require adjustment. What is the probability that in a sample of 20 window frames, at least one will need adjustment?

64.15% probability that in a sample of 20 window frames, at least one will need adjustment

Step-by-step explanation:

For each window frame, there are only two possible outcomes. Either it needs adjustment, or it does not. The probability of a window frame needing adjustment is independent of other window frames. So 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.

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

And p is the probability of X happening.

5% of the production will have some type of minor defect that will require adjustment.

This means that

What is the probability that in a sample of 20 window frames, at least one will need adjustment?

This is when n = 20.

We know the either no windows need adjustment, or at least one does. The sum of the probabilities of these events is decimal 1. So

In which

64.15% probability that in a sample of 20 window frames, at least one will need adjustment

answer:

it is general formula that is used to write value to its excet position

a11 a12 a13

a21 a22 a23

it is the example of order 2 by 3 matrix in this

a11 means value occur in 1 st row 1st column

a12 means value occurs in 1st row 2nd column

a13 means value occur in 1st row and 3rd column

a21 means value occurs in 2nd row and 1st column

a22 means value occurs in 2nd row and in 2nd column

a23 means value occurs in 2nd row and in 3rd column