Answer:
28.6, that is, about 29 are expected to be defective
Step-by-step explanation:
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 other betteries. So the binomial probability distribution is used to solve this question.
Binomial probability distribution
Probability of exactly x sucesses on n repeated trials, with p probability.
The expected value of the binomial distribution is:
[tex]E(X) = np[/tex]
The probability that a battery is defective is 1/14.
This means that [tex]p = \frac{1}{14}[/tex]
400 batteries.
This means that [tex]n = 400[/tex]
How many are expected to be defective?
[tex]E(X) = np = 400*\frac{1}{14} = 28.6[/tex]
28.6, that is, about 29 are expected to be defective
