Question

A light bulb has a lifetime that is exponential with a mean of 200 days. When it burns out a janitor replaces it immediately. In addition there is a handyman who comes at times of a Poisson process at rate .01 and replaces the bulb as "preventive maintenance." (a) How often is the bulb replaced? (b) In the long run what fraction of the replacements are due to failure?

Answer 1

Answer:

(a) The number of bulbs often replaces is 66.67.

(b) The fraction of the replacements that are due to failure, in the long run, is $\frac{1}{3}$.

Step-by-step explanation:

Let X = lifetime of a bulb and Y = time after which the bulb is replaced.

It is provided that X follows Exponential distribution with mean lifetime of a bulb is, 200 days.

And the rate at which the bulb is replaced is, 0.01 also following an Exponential distribution.

(a)

A bulb is replaced only after it burns out or a handyman comes at times of a Poisson process and replaces it.

Then min (X, Y) follows an Exponential distribution with parameter $(\frac{1}{200}+0.01)$.

The mean of an Exponential distribution with parameter θ is:

$Mean=\frac{1}{\theta}$

Compute the mean of min (X, Y) as follows:

$Mean =\frac{1}{(\frac{1}{200}+0.01)} =\frac{1}{0.015}= 66.67$

Thus, the number of bulbs often replaces is 66.67.

(b)

Compute the probability of the event (X < Y) as follows:

$P(X$

Thus, the fraction of the replacements that are due to failure, in the long run, is $\frac{1}{3}$.