Using the binomial distribution, it is found that about 75 batteries each day are defective.
For each battery, there are only two possible outcomes, either it is defective, or it is not. The probability of a battery being defective is independent of any other battery, hence the <em>binomial distribution</em> is used to solve this question.
<h3>What is the binomial probability distribution?</h3>It is the probability of exactly <u>x successes on n repeated trials, with p probability</u> of a success on each trial.
The expected value of the binomial distribution is:
In this problem:
- 3 out of 20 batteries are defective, hence p = 3/20 = 0.15.
- Each day, 500 batteries are produced, hence n = 500.
Then, the expected number of defective batteries in a day is given by:
E(X) = np = 500(0.15) = 75.
More can be learned about the binomial distribution at brainly.com/question/14424710
