Respuesta :
Answer:
The probability that a skid of 50 DVD players will contain at least 3 defective units is 0.9487.
Step-by-step explanation:
We are given that Avery and Bradly work at a large electronics manufacturer that produces DVD players. The defective rate on the assembly line has gone up 12% and a skid of 50 DVD players has been selected by the manager.
Let X = Number of defective units of DVD players
The above situation can be represented through the binomial distribution;
[tex]P(X = r) = \binom{n}{r}\times p^{2} \times (1-p)^{n-r}; x= 0,1,2,3,....[/tex]
where, n = number of samples (trials) taken = 50 DVD players
r = number of success = at least 3 defective units
p = probability of success which in our question is the probability
of defective rate, i.e; p = 12%
So, X ~ Binom(n = 50, p = 0.12)
Now, the probability that a skid of 50 DVD players will contain at least 3 defective units is given by = P(X [tex]\geq[/tex] 3)
P(X [tex]\geq[/tex] 3) = 1 - P(X = 0) - P(X = 1) - P(X = 2)
= [tex]1 - [ \binom{50}{0}\times 0.12^{0} \times (1-0.12)^{50-0}]-[ \binom{50}{1}\times 0.12^{1} \times (1-0.12)^{50-1}]-[ \binom{50}{2}\times 0.12^{2} \times (1-0.12)^{50-2}][/tex]
= [tex]1 - [ 1 \times 1 \times 0.88^{50}]-[50 \times 0.12^{1} \times 0.88^{49}]-[ 1225 \times 0.12^{2} \times 0.88^{48}][/tex]
= 0.9487
Hence, the probability that a skid of 50 DVD players will contain at least 3 defective units is 0.9487.
