If 500 fuses are produced each day, how many can we expect to find each day that are defective ?

Question

Consider the following binomial experiment: The probability that a fuse produced by a certain company will be defective is 37⁄100.

Answer 1

185

E(d) = np = 500(37/100) = 5(37) = 185 = expected defects

n = number potentially defective

p = probability of a defect

np = average = Expected value

Answer 2

Hi Terah,

The expected value or E(X)= np.In this case E(X)=500(37/100) = 185