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11. Avery and Bradley work at a large electronics manufacturer that produces DVD players. The
defective rate on the assembly line has gone up to 12% and the manager wants to know the
probability that a skid of 50 DVD players will contain at least 3 defective units.
a) Help Avery use the binomial distribution P(x)=, C.pqrs to answer this question

Answer :

raphealnwobi

Answer:

0.9487

Step-by-step explanation:

a) The probability of having a defective product = p = 12% = 0.12

The probability of not having a defective product = q = 1 - p = 1 - 0.12 = 0.88

The number of DVD players = n = 50

X is the number of defective products.

The  probability that a skid of 50 DVD players will contain at least 3 defective units = P(X ≥ 3) = 1 - P(X ≤ 2)

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

Using binomial:

P(X = 0) = [tex]C(n,x)p^xq^{n-x}=C(50,0)p^0q^{50-0}=\frac{50!}{(50-0)!0!}*0.12^0*0.88^{50-0}=0.0017[/tex]

P(X = 1) = [tex]C(n,x)p^xq^{n-x}=C(50,1)p^1q^{50-1}=\frac{50!}{(50-1)!1!}*0.12^1*0.88^{50-1}=0.0114[/tex]

P(X = 2) = [tex]C(n,x)p^xq^{n-x}=C(50,2)p^2q^{50-2}=\frac{50!}{(50-2)!2!}*0.12^2*0.88^{50-2}=0.0382[/tex]

P(X ≤ 2)= 0.0017 + 0.0114 + 0.0382 = 0.0513

P(X ≥ 3) = 1 - P(X ≤ 2) = 1 - 0.0513 = 0.9487 = 94.87%

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