Question

3. A manufacturer knows that, on average, 3% of items manufactured will have defects. Use the normal approximation to the binomial distribution to determine the probability that among 200 items, (a) at the most 5 will be defective; (b) anywhere from 4 to 7 will be defective 4. Page 140, #5.26 5. Page 178 #5.109

Answer 1

N = 200 p = 3% = 3/100 N = 200 (very large) Let X be the number of defective item out of N E (x) = lambda - 200 times 3/100 = 6 (a) p (x lessthanequalto 5) = e^-lambda summation^5_k = 0 lambda^k/k! = e^-6 summation^5_k = 0 6^k/k! (b) p (4 lessthanequalto x lessthanequalto 7) = e^-6 [6^4/4! + 6^5/5! + 6^6/6! + 6^7/7!]