The complete question is
"Compute the expectation and variance of the number of successes in n independent trials, each of which results in a success with probability p. Is independence necessary? "
Are any of my answers correct?
Xi = {1 If success} {0 if failure}
- For the expectation part of the question, I got that
E[X] = E[X1] + ... + E[Xn] = (1/n)n = 1
- For the variance part of the question, I got that
Var(x) = 1^2(p{success}) + 0^2(p{success}) - 1^2
- Is independence Necessary? I'm not sure at all at what this question is asking about.
Thank You
Your definition of $X_i$ is well chosen. By linearity of the expectation operator, $$ E\left(\sum_{i=1}^n X_i\right)= \sum_{i=1}^n E(X_i)= n E(X_1)$$ since the $X_i$ are all identically distributed. You made a mistake when calculating the expected value: $$E(X_1)=1\cdot P(X_1=1) +0\cdot P(X_1=0)=p$$ Hence, expected number of successes is $np$ which (I think) also makes intuitive sense. Assuming independence, we have that $$Var\left(\sum_{i=1}^n X_i\right)= \sum_{i=1}^n Var(X_i)= n Var(X_1)$$ Try to find $Var(X_1)$ by yourself.
For the independence part: I personally don't know what they mean by asking if independence is "necessary". If they are independent, then the variance can be calculated as above, if they are not independent, one would have to take into account covariances in the variance formula. Note that, no matter if independent or not, the expected value is linear, hence the expected number of successes is $np$.
Note that the number of successes is in $n$ independent trials with same success probability is a Binomial distribution, more about that here: https://en.wikipedia.org/wiki/Binomial_distribution