Design an op amp circuit to average the input - id6190985, divyaravinder8148

Ответ:

xxtonixwilsonxx

26.11.2020

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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