Question

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume that the probability that a wafer passes the test is 0.7 and that wafers are independent. Determine the cumulative distribution function of the number of wafers from a lot that pass the test at specified values. Round your answers to three decimal places (e.g. 98.765). F(O) = P(X=0) = .027 F(1) = P(X= 1) = i .189 F(2) = P(X + 2) = 441 F(3) = P(X + 3) = .343

Answer 1

det X that pass the be the no. of wafers - test Xu Beinomial (3, 0.7) 199 P(O) = P1x<0) = P(X= 0) = 360 (0.4)º(0-3)*= 0.027 b) f(1) = P(x 1) = P(x=0) +P(X=1) = °C10.73°0.3)+ 36, 10.7) (0.3)2 = 0.027 +0.189 = 0.216 1 F(2) = P(X<2) = F(1) + P(X=2) = 0.216 + 3C2 (0.7) 70.3) -0.216 +0.441 = 0.657 D F13) = P(X33) = F(2) + P(x=3) = 0.657 + 303 (0.7)3 = 0.657 +0.343