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g100num [7]
3 years ago
15

In a semiconductor manufacturing process, three wafers from a lot are tested. Each wafer is classified as pass or fail. Assume t

hat the probability that a wafer passes the test is 0.6 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).
Engineering
1 answer:
gladu [14]3 years ago
5 0

Answer:

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

Explanation:

Each wafer is classified as pass or fail.

The wafers are independent.

Then, we can modelate X : ''Number of wafers that pass the test'' as a Binomial random variable.

X ~ Bi(n,p)

Where n = 3 and p = 0.6 is the success probability

The probatility function is given by :

P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}

Where nCx is the combinatorial number

nCx=\frac{n!}{x!(n-x)!}

Let's calculate f(x) :

f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064

f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288

f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432

f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216

For the cumulative distribution function that we are looking for :

P(X\leq x)=F(x)

F(0)=f(0)\\F(1)=f(0)+f(1)\\F(2)=f(0)+f(1)+f(2)\\F(3)=f(0)+f(1)+f(2)+f(3)=1

F(0)=0.064\\F(1)=0.064+0.288=0.352\\F(2)=0.064+0.288+0.432=0.784\\F(3)=0.064+0.288+0.432+0.216=1

The cumulative distribution function for X is :

F(x) = 0           ;  x < 0

         0.064   ;  0 ≤ x < 1

         0.352   ;  1 ≤ x < 2

         0.784   ;  2 ≤ x < 3

            1        ;    x ≥ 3

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