Consider three lightbulbs each of which has a lifetime that is an independent exponential random variable with parameter λ=1. The variance of the time until all three burn out is:
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Recall that the variance of an exponential with parameter λ is 1/λ2.
1/9 +1/4 + 1 = 1.361111
Well, you know what they say about lightbulbs, they never seem to last! So let's shed some light on this problem.
The time until each lightbulb burns out follows an exponential distribution with parameter λ = 1. Now, because the lifetimes of the lightbulbs are independent, we can just add up their lifetimes to find the time until all three burn out.
Now, the variance of an exponential distribution with parameter λ is 1/λ^2. So for each lightbulb, the variance of its lifetime is 1/1^2 = 1. Since we have three lightbulbs, and their lifetimes are independent, we can just add up their variances to find the variance of the time until all three burn out.
Since variances add up for independent variables, the variance of the time until all three lightbulbs burn out is 1 + 1 + 1 = 3.
So there you have it, the variance of the time until all three lightbulbs burn out is 3. Now let's hope they don't leave you in the dark for too long!
To find the variance of the time until all three lightbulbs burn out, we can use the fact that the sum of independent exponential random variables follows a gamma distribution. In this case, since we have three exponential random variables, the sum follows a gamma distribution with shape parameter k = 3 and scale parameter θ = 1/λ = 1/1 = 1.
The variance of a gamma distribution with shape parameter k and scale parameter θ is given by the formula Var(X) = k * θ^2.
Substituting the values of k and θ, we have Var(X) = 3 * (1)^2 = 3.
Therefore, the variance of the time until all three lightbulbs burn out is 3.