A manufacturing machine has a 50% defect rate. If 102 items are chosen at random, answer the following.

Question

What is the probability that exactly 47 of them are defective?

What is the probability that less than 47 of them are defective?

What is the probability that more than 47 of them are defective?

What is the probability that exactly 57 of them are defective?

What is the probability that at least 57 of them are defective?

What is the probability that at most 57 of them are defective

Answer 1

use normal approximation to binomial distribution and apply continuity factor (+ or - 0.5) because approximating discrete distribution (binomial) with a continuous one (normal).

z = (x - np)/(sqrt(npq), where = 102, p = 0.5 and q = 0.5 and x = number defective:

z = (x - 51)/5.05

P(x = 47) = P(46.5 < x < 47.5) = P( (46.5 - 51)/5.05 < z < (47.5 - 51)/5.05) = (-0.89 < z < -0.69) = P(z < -0.69) - P(z < -0.89) = 0.2451 - 0.1867 = 0.0674

P(x < 47) = P(x < 46.5) = P(z < -0.89) = 0.1867

P(x > 47) = 1 - P(x = 47) - P(x < 47) = 1 - 0.1867 - 0.0674 = 0.746

P(x = 57) = P(56.5 < x < 57.5) = P((56.5 - 51)/5.05 < z < (57.5 - 51)/5.05) = P(1.09 < z < 1.29) = P(z < 1.29) - P(z < 1.09) = 0.9015 - 0.8621 = 0.0394

P(x >= 57) = P(x > 56.5) = P(z > (56.5 - 51)/5.05) = P(z > 1.09) = 1 - P(z < 1.09) = 1 - 0.8621 = 0.1379

P(x <= 57) = P(x < 57.5) = P(z < 1.29) = 0.9015