During a manufacturing process 15 units are randomly selected each day from the production line to check the percent defective. from the historical information it is known that the probability of a defective unit is 0.05. any time that two or more defectives are found in the sample of 15, the process is stopped. this procedure is used to provide a signal in case the probability of a defective is increased. what is the proability that on any given day the production process will be stopped?

There is a 17.09% proability that on any given day the production process will be stopped

Step-by-step explanation:

The production will be stopped if there are two or more defective units, so we have to find .

This is . We know that either we have less than 2 defecive units, or we have at least 2 defective units. The sum of the probabilities is decimal 1. So

So, we have to find P(X < 2), that is:

Since either an unit is defective or it is not, that is, there are ony two possible outcomes, P(X = 0) and P(X = 1) are binomial probabilities.

Binomial probability

Th 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 combinatios of x objects from a set of n elements, given by the following formula.

And is the probability of X happening.

So:

The probability of a defective unit is 0.05, so

The size of the sample is 15, so

There is a 17.09% proability that on any given day the production process will be stopped

1. 34

2. 268

3. x=14.4

step-by-step explanation:

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