a) Probability contains at least 3 = 1 - probability contains less than three = 1 - (probability contains 0 or 1 or 2)

Binomial distribution formula k successes in n tries where p = probability of success C(n,k) p^k (1-p)^(n-k)

Use binomial formula for k=0, 1, 2 and p = 0.12 and sum probabilities. (answer is 0.9487)

b) want to find P(x >= 3) where x is number of defective units.

= 1 - P(x < 3)

for normal approximation to binomial

mean = np = 50 * 0.12 = 6

sd = sqrt(npq) = sqrt (50 * 0.12 * 0.88) = 2.298

z = x - mean / sd = x - 6 / 2.298

P(x < 3) = P(z < 3 - 6 / 2.298) = P( z < -1.31) = 0.0951

1 - P(x < 3) = 1 - 0.0951 = 0.9049

Zaina R.

also wouldnt 12% represent the probability of failure and 88% would be the success?08/25/20

Robert Z.

Sounds crazy I know, but since this problem is focusing on failures, finding them is considered "success." The problem could have asked for the probability that there are 47 or fewer non-defective units. The answers would come out the same.08/25/20

Robert Z.

Also, when doing the normal approximation to the discrete binomial distribution, all the continuous values from 1.5 to 2.5 represent the 2's and the values from 2.5 to 3.5 represent the 3's. This means that the normal approximation should be written P(x < 3) = P(z < 2.5 - 6 / 2.298) = P( z < -1.523) = 0.0639 1-0.0639 = .9361 This is much closer to the binomial result.08/25/20

Vinu P.

@Zaina. I have the same question too. Basically the manager wants to know what the probability of defective units are. 12% is the defective rate and is also used as a success. Failure would be 88% because that is the non-defective rate. The question asks for defective units so you would use 12 % as the success to solve the question. If they asked for units that were not defective then the success would be 88%.08/25/20

Zaina R.

how do I show 0,1, and 2 in the binomial formula08/25/20