Question

Dashboard / My courses / PROBABILITY AND ENGINEERING STATISTICS-1194-mota / Chapter Question Hot yet owered Morked out of 500 Suppose that when a machine is adjusted properly, 87% of the items produced by it are of high quality and the others are of medium quality Suppose however, that the machine is improperly adjusted during 14% of the time and that under these conditions 13% of the items produced by it are of high quality and the remaining are of medium quality. If an item is selected at random and found to be of medium quarty, what is the probability that it was produced while the machine was adjusted property The answer should be a number rounded to five decimal places. don't use symbols such as $] Hog question Next page

Answer 1

= 0.86 P (high quality / properly adjusted) 0.87 - P (H/P) P(medium quality /properly adjusted). 0-13 - P(M/P) P (Improperly adjusted ) = P(P) = 0.14 P (P) = 1-0.14 P(H/F) = 0.13 P(M/F) = 0.87 P (properly adjusted/ medium quality) . P(P/M) - P(M/P) P(P) P(M/P) P(1) + P (M/F) (F) 0.13 X 0.86 0.13 X 0.86 + 0.878 0.14 o.1118 0.1118+ 0.1218 0.1118 0.2336 0.47859