Confidence interval for mean Let's say we draw $49$ numbers from $1,\ldots,100$ without returning them back. Then we use the arithmetic mean from the sample.
$$M=\frac{1}{49}\sum_{i=1}^{49}X_i$$
They gave me the hint that $M$ is roughly normal distributed despite the dependencies in the draws. Now I have to determine an symmetric intervall $J$ around the mean with $\mathbb{P}(M\in J)\approx0.95$
My idea was to use the variance and covariance to solve this problem.
$$Var\left(\frac{1}{49}\left(X_1+\ldots+X_{49}\right)\right)$$
$$=\frac{1}{49^2}\left(49\cdot Var(X_1)+49\cdot48Cov(X_1,X_2\right))$$
But I'm not sure how this works
 A: The expected value of the mean $M$ is
$$
\mu = \frac{1}{49} \sum_{i=1}^{49} \Bbb E X_i \, ,
$$
and its variance is
$$
\sigma^2 = \frac{1}{49^2} \left( \sum_{i=1}^{49} \text{var} X_i + 2 \sum_{1\leq i<j\leq 49} \text{cov} (X_i,X_j) \right) .
$$
Following the indication, we may assume that $M \sim \mathcal{N}(\mu,\sigma^2)$, or equivalently $\tfrac{M - \mu}{\sigma} \sim \mathcal{N}(0,1)$. Now, we are looking for $J = \left[\mu-\varepsilon,\mu+\varepsilon\right]$ such that $\Bbb P (M \in J) = 0.95$, or equivalently,
$$
\Bbb P \left(-\tfrac{\varepsilon}{\sigma} \leq \tfrac{M-\mu}{\sigma} \leq \tfrac{\varepsilon}{\sigma}\right) = 0.95 \, .
$$
A sketch of the standard normal's p.d.f. shows that
$\Phi(\frac{\varepsilon}{\sigma}) = 0.975$, where $\Phi$ denotes the c.d.f. of the standard normal. Therefore, $\varepsilon = \sigma\, \Phi^{-1}(0.975) \approx 1.96\, \sigma$. A confidence interval of this type for $M$ is therefore
$$
J \approx \left[\mu-1.96\, \sigma,\mu+1.96\, \sigma\right]\, .
$$
A: To determine $\mathrm{cov}(x_1,x_2)$, there is a trick. We have
$$\mathrm{var}(x_1+x_2+\cdots+x_{100})=0=100\,\mathrm{var}(x_1)+100\times 99\,\mathrm{cov}(x_1,x_2),$$
because when all numbers are drawn, the result is certain, and because every number is treated on equal footing in the random procedure, $\mathrm{var}(x_1)=\mathrm{var}(x_2)=\ldots=\mathrm{var}(x_{100})$, and $\mathrm{cov}(x_1,x_2)=$ $\mathrm{cov}(x_1,x_3)=\ldots=\mathrm{cov}(x_{99},x_{100})$. Since we know 
$$\mathrm{var}(x_1)=\frac{1}{100}(1^2+2^2+\cdots+100^2)-\left(50\frac{1}{2}\right)^2=833\frac{1}{4},$$
from the first equation we have
$$\mathrm{cov}(x_1,x_2)=-\frac{1}{99}\mathrm{var}(x_1)=-8\frac{5}{12}.$$
Then you can determine the variance of the partial sum $\,x_1+x_2+\cdots+x_{49\,}$ and the confidence interval accordingly.
