Using the binomial distribution, it is found that there is a:
a) 0.3526 = 35.26% probability of rejecting a lot that is 3% defective.
b) 0.4295 = 42.95% probability of accepting a lot that is 4% defective.
For each device, there are only two possible outcomes, either it is defective, or it is not. The probability of a device being defective is independent of any other device, hence the binomial distribution is used to solve this question.
Binomial probability distribution
The parameters are:
- x is the number of successes.
- n is the number of trials.
- p is the probability of a success on a single trial.
In this problem, the sample has 100 units, hence .
Item a:
3% of the pieces are defective, hence .
The probability is:
In which:
Hence:
Then:
0.3526 = 35.26% probability of rejecting a lot that is 3% defective.
Item b:
4% of the pieces are defective, hence .
Lot is accepted if <u>less than 4 units are defective</u>, hence:
Then:
0.4295 = 42.95% probability of accepting a lot that is 4% defective.
A similar problem is given at brainly.com/question/24863377