Respuesta :
Answer:
A) probability of failure in next 100 hours given that it has been tested for 500 hours without failure is 0.181
B) probability that exactly two have the metabolic defect is 0.03
Step-by-step explanation:
Part A)
Let X be a exponentially random variable with mean = μ = 500 hrs
For exponential distribution:
[tex]p.d.f = f(x) = \lambda e^{-\lambda x}\\c.d.f = F(x) = 1 - e^{-\lambda x}\\x\geq 0[/tex]
λ = 1/μ
λ = 0.002
We have to find the probability of failure in the next 100 hours given that assembly has been tested for 500 hours without a failure.
Using memory less property of exponential distribution:
[tex]P(X<500 + 100|X>500) = P (X<100)\\ P (X<100)= F(100)[/tex]
using
[tex]F(x) = 1 - e^{-\lambda x}\\ \lambda =.002\\x=100\\F(x) = 1- e^{-(.002)(100)}\\F(x) = 1-.8187\\F(x) = 0.181[/tex]
Part B)
Chances of occurrence of metabolic defect = 5%
P(C) = .05
No. of randomly selected infants = n =6
We have to find the probability that exactly two have the metabolic defect
⇒x = 2
Using binomial probability density function:
P = [tex]P=\left[\begin{array}{ccc}n\\x\end{array}\right] p^{x} (1-p) ^{n-x}\\\\=\frac{n!}{x!(n-x)!} p^{x} (1-p) ^{n-x}\\=\frac{6!}{2!4!}(.05)^{2}(.95)^{4}\\= 0.03\\[/tex]
probability that exactly two have the metabolic defect is 0.03
