Question

Can this question be solved Confidence Interval Formulas at all? Every answer seems to use P -Values and my course has not covered that yet. Historically, a production line produces 6% defective items. The production supervisor takes a sample of 100 items frequently and if he finds 8 or more defective products, he stops the line to make adjustments. The probability that a random sample of 100 would lead to the stoppage of the production line is: (a) 0.2995 (b) 0.3893 (c) 0.2005 (d) 0.4584 (e) 0.7995

Answer 1

Given that: n 100 and p 690 0.06 Calculate the mean and standard deviation -100 x0.06 100x0.06(1-0.06) - 2.37 Let X denote the defective products in stoppage ofthe production line

a) Calculate the probability that 8 or more defective products. P(X28)-1- P(X <8) ơ 2.37 = 1-P(z 0.84) =1-1-NORMSDIST(0.84) = 1-0.799546 (Use MS Excel) 0.200454 0.2005 (Round to 4 decimal place) Hence, the required answer is 0.2005 The correct option is: C