Question

The lifetime of a mechanical assembly in a vibration test is exponentially distributed with a mean of 500 hours. If an assembly has been on test for 500 hours without a failure, what is the probability of a failure in the next 100 hours?

b) A metabolic defect occurs in approximately 5% of the infants born at a hospital. Six infants born at the hospital are selected at random. What is the probability that exactly two have the metabolic defect?

Answer 1

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

Explanation:

Part A)

Let X be a exponentially random variable with mean = μ = 500 hrs

For exponential distribution:

`p.d.f = f(x) = \lambda e^(-\lambda x)\\c.d.f = F(x) = 1 - e^(-\lambda x)\\x\geq 0`

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:

`P(X<500 + 100|X>500) = P (X<100)\\ P (X<100)= F(100)`

using

`F(x) = 1 - e^(-\lambda x)\\ \lambda =.002\\x=100\\F(x) = 1- e^(-(.002)(100))\\F(x) = 1-.8187\\F(x) = 0.181`

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 =
`P=\left[\begin{array}{ccc}n\\x\end{array}\right] p^(x) (1-p) ^(n-x)\\\\=(n!)/(x!(n-x)!) p^(x) (1-p) ^(n-x)\\=(6!)/(2!4!)(.05)^(2)(.95)^(4)\\= 0.03\\`

probability that exactly two have the metabolic defect is 0.03