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 [tex]\frac{1}{3}[/tex].
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 [tex](\frac{1}{200}+0.01)[/tex].
The mean of an Exponential distribution with parameter θ is:
[tex]Mean=\frac{1}{\theta}[/tex]
Compute the mean of min (X, Y) as follows:
[tex]Mean =\frac{1}{(\frac{1}{200}+0.01)} =\frac{1}{0.015}= 66.67[/tex]
Thus, the number of bulbs often replaces is 66.67.
(b)
Compute the probability of the event (X < Y) as follows:
[tex]P(X<Y)=\frac{0.005}{0.015} =\frac{1}{3}[/tex]
Thus, the fraction of the replacements that are due to failure, in the long run, is [tex]\frac{1}{3}[/tex].
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 .
Step-by-step explanation:
