In a semiconductor manufacturing process, three wafers from a lot are tested. Ea

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 0.8 and that wafers are independent. Sample Space: S = {PPP, PPF, PFP, FPP, PFF, FPF, FFP, FFF} Let X denote the number of wafers that pass (a) Fill out the following table: (b) The probability mass function of X is partially completed in the table below. Fill in the blanks. (c) What is the cumulative distribution function (cdf) of X. (d) What is he probability that exactly two wafers pass? (e) What is the probability that no more than 3 wafers pass?

Explanation / Answer

a) The success probability is given as 0.8 and failure probability is given as 1 - 0.8 = 0.2

Therefore the distribution table is filled as:

Here the fourth column represents the required PDF

b) The probability mass function of X is computed as:

c) Cumulative probability ( CDF ) for X is computed as:

The last row represents the required CDF

d) Probability that exactly 2 wafers pass is given as: P(X =2 ) = 0.096

e) Probability that no more than 3 wafers pass is 1 as there are only 3 wafers.

Wafer 1 Wafer 2 Wafer 3 x f(x) - PDF F(x) - CDF P P P 3 0.83 = 0.512 0.512 P P F 2 0.82*0.2 = 0.128 0.512 + 0.128 = 0.64 P F P 2 0.82*0.2 = 0.128 0.64 + 0.128 = 0.768 F P P 2 0.82*0.2 = 0.128 0.768 + 0.128 = 0.896 P F F 1 0.8*0.22 = 0.032 0.896 + 0.032 = 0.928 F F P 1 0.8*0.22 = 0.032 0.928 + 0.032 = 0.96 F P F 1 0.8*0.22 = 0.032 0.96 + 0.032 = 0.992 F F F 0 0.23 = 0.008 0.992 + 0.008 = 1

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.