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.

08/25/20

Robert Z.

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