Find expected value of exponential distribution

Question

I need to find the expected value of an exponential distribution given two means, my problem is i am not sure if i am rigth.
The average life time of an AmpOp can be modeled as an exponential distribution.
If 10% of them has average life time of 20000 hours and 90% has average time life of 50000 hours, find the proportion that stops to work before 60000 hours.

My question. Can i assume that 10% and 90% are independent random variables?
If so i will have:
$\lambda_1 $ for 10% so that: $P(X \leq t)=1- e^{-\lambda_1 t}$
$\lambda_2 $ for 90% so that: $P(Y \leq t)=1-e^{-\lambda_2 t}$
$P(X \leq t,Y \leq t)=1-e^{-\lambda_1 t}-e^{-\lambda_2 t}+e^{-(\lambda_1+\lambda_2) t}$

Or i can calculate $E(X)=0.1*20000+0.9*50000=47000$ and $\lambda =1/E(X) \ $?

Answer 1

The question does not ask for an expected value

so you would do better to calculate:

• the proportion of the $20000$-hour-average items which stop working before $60000$ hours
• the proportion of the $50000$-hour-average items which stop working before $60000$ hours

and then combine these two figures in a weighted sum

Answer 2

You are told that for the $10\%$, the failure rate is $1/20000$ per hour, and for the $90\%$ the rate is $1/50000$ per hour.

You seek the proportion of the population that fails before a given time.   By the Law of Large Numbers, that is trends to the probability that an individual fails in that time.   By the Law of Total Probability, that will be a weighted probability over each group.

$$\mathsf P(X\leq 60000) = 0.10 \mathsf P(X_{10\%}\leq 60000)+0.90\mathsf P(X_{90\%}\leq 60000)$$