# Exponential distribution

• Apr 3rd 2011, 06:14 AM
tsang
Exponential distribution
Hi, can anyone please help me with this question? I really don't know how to do it. Thanks a lot. I'm a bit confused with the situation.

The time to failure of a new type of light bulb is thought to have a exponential distribution.

(a)The reliability of a component is defined as the probability that the component will not have failed by a specified time. If the reliability of this type of light bulb at 10.5 weeks if 0.9, find the reliability at 10 weeks.
(b)One hundred light bulbs of this type are put in a new shop. All light bulbs that have failed are replaced at 20 week intervals and none are replaced at other times. If R is the number of light bulbs that have to be replaced at the end of the first interval, find the mean and variance of R. Explain why this result holds for any such interval, and not just the first.

For Part (a), I did try to use fomula, so I tried to solve $0.9=e^{-\alpha*10.5}$, then $\alpha=0.01003$, then I did $Pr(T=10)=e^-0.01003*10=0.9045$, so should my answer be 0.9045? Did I use right way?
I really don't know how to do Part (b), please help me with it. Thanks a lot.
• Apr 3rd 2011, 10:31 AM
Sambit
Quote:

Originally Posted by tsang
For Part (a), I did try to use fomula, so I tried to solve $0.9=e^{-\alpha*10.5}$, then $\alpha=0.01003$, then I did $Pr(T=10)=e^-0.01003*10=0.9045$, so should my answer be 0.9045? Did I use right way?

Your answer is correct but 0.9045 is NOT P(T=10); rather it is P(T>10).
• Apr 3rd 2011, 07:53 PM
tsang
Quote:

Originally Posted by Sambit
Your answer is correct but 0.9045 is NOT P(T=10); rather it is P(T>10).

Thanks for your help. Just wonder, could you please help me with part (b), it is quite hard for me. Thanks a lot.
• Apr 3rd 2011, 11:05 PM
Sambit
Note that $\sum_{i=1}^{100}T_i \sim exp(100*0.01003)$. Then what will be the mean and variance for $\sum_{i=1}^{100}T_i$ ? Calculate this first. Then you need to calculate mean and variance of $100\sum_{i=1}^{100}T_i$. Give it a try.
• Apr 4th 2011, 01:27 PM
theodds
Quote:

Originally Posted by Sambit
Note that $\sum_{i=1}^{100}T_i \sim exp(100*0.01003)$. Then what will be the mean and variance for $\sum_{i=1}^{100}T_i$ ? Calculate this first. Then you need to calculate mean and variance of $100\sum_{i=1}^{100}T_i$. Give it a try.

The sum of independent exponentials is a gamma, not an exponential.

For part (b), when explaining why the result holds for time intervals other than the first, it is probably sufficient to just appeal to the fact that the exponential distribution is memoryless. At any rate, the distribution of R looks to me to be binomial, with success probability equal to the probability that a component burns out before the 20 weeks are up, and 100 trials. So, just apply what you know about the binomial distribution to calculate the mean and variance of R.
• Apr 4th 2011, 04:44 PM
tsang
I will try again, thanks a lot. If I still can't get it, I will least show what I can do at this stage. Thanks again.