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

56.6

step-by-step explanation:

the ratio of the perimeters is equal to the scale.

the ratio of the areas is equal to the square of the scale.

34 / 20 = k

a / 19.6 = k²

substituting:

a / 19.6 = (34 / 20)²

a = 19.6 (34 / 20)²

a ≈ 56.6

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step-by-step explanation: