Answer:
F(x) = 0 Â Â Â Â Â ; Â 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 :
[tex]P(X=x)=f(x)=nCx.p^{x}.(1-p)^{n-x}[/tex]
Where [tex]nCx[/tex] is the combinatorial number
[tex]nCx=\frac{n!}{x!(n-x)!}[/tex]
Let's calculate f(x) :
[tex]f(0)=3C0.(0.6^{0}).(0.4^{3})=0.4^{3}=0.064[/tex]
[tex]f(1)=3C1.(0.6^{1}).(0.4^{2})=0.288[/tex]
[tex]f(2)=3C2.(0.6^{2}).(0.4^{1})=0.432[/tex]
[tex]f(3)=3C3.(0.6^{3}).(0.4^{0})=0.6^{3}=0.216[/tex]
For the cumulative distribution function that we are looking for :
[tex]P(X\leq x)=F(x)[/tex]
[tex]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[/tex]
[tex]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[/tex]
The cumulative distribution function for X is :
F(x) = 0 Â Â Â Â Â ; Â x < 0
     0.064  ;  0 ≤ x < 1
     0.352  ;  1 ≤ x < 2
     0.784  ;  2 ≤ x < 3
      1     ;   x ≥ 3
The answer is "0.064, 0.352,0.784, and 0.999".
[tex]\to P(x) = \binom{n}{x} p^x q^{n-x}\\\\\to n=3 \\\\\to p=0.6 \\\\\to q=1-p=1-0.6=0.4 \\\\[/tex]
Calculating the distribution function:
[tex]\to F(0) = P(x \leq 0) = P(x=0) = \binom{3}{0} 0.6^0\ 0.4^3= 0.064 \\\\\to F(1) = P(x \leq 0) = P(x=0)+p(x=1) = 0.064+3(0.6)(0.16)= 0.64+0.288= 0.352\\\\ \to F(2) = P(x \leq 2) = P(x=0)+p(x=1)+p(x=2) = 0.064+0.288 + \binom{3}{2}(0.6)^2(0.4)= 0.784\\\\\to F(3) = P(x \leq 3) = P(x=0)+p(x=1)+p(x=2) +p(x=3)= 0.064+0.288 + 0.432+\binom{3}{3}(0.6)^3= 0.999\\\\[/tex]
Therefore, the answer is "0.064, 0.352,0.784, and 0.999".
Find out more information about the distribution function here:
brainly.com/question/26062265
