it seems the variance of exponetial is 1/labda^2 and of binomial distribution is np(1-p). But isn't variance E(X^2) - E(X)^2 ? and dont i need to take laplace transform and take derivative to find the moments?
Q : Iwana Passe is taking a quiz with 12 questions. The amount of time she spends answering question i is Ti, and is exponentially distributed with E[Ti] = 1/3 hour. The amount of time she spends on any particular question is independent of the amount of time she spends on any other question. Once she finishes answering a question, she immediately begins answering the next question.
Let N be the total number of questions she answers correctly.
Let X be the total amount of time she spends on questions that she answers correctly.
For parts (a) and (b), suppose we know she has probability 2/3
of getting any particular quiz question correct, independently of her performance on any other quiz question.
(a) Find the expectation and variance of X.
The solution is:
Let X = T1 + T2 + ... + TN where N is a binomial with parameters p = 2/3
and n = 12.
E[Ti] = 1/3, thus, Ti is an exponential with rate = 3, so fTi(t) = 3e^(−3t) with t ≥ 0.
E[X] = E[Ti]E[N] = 1/3 ∗ 12 ∗ 2/3 = 8/3
var(X) = var(Ti)E[N] + (E[Ti])^(2)var(N) = 1/9 ∗ 12 ∗ 2/3 + 1/9 ∗ 12 ∗ 2/3 ∗ 1/3 = 32/27
The E[Ti] =1/3 i get since all questions takes 1/3 time to finish and that is the mean. I get that the E[X] is a combination of the average of time Ti and N questions answered correctly.
1. How come E[N] is just 12*2/3? Is that because we know the expected value will always be 2/3 * N since she answers that many correct?
2. I am not sure how he came to the solution of the variance there - if he takes var(Ti) why is it 1/9 and how come Var(N) = 12* 2/3 * 1/3 ? Isn't σ2=E(X^2) - E(X)^2 ?
I think they are combining 2 variances but how do you get E(X^2) on a constant?
some clarifications is appreciated!