This will seem like a very simple question to many of you; but I cannot understand part of the solution, so to give context I have had to unfortunately resort to typing the whole thing out, apologies.
Start of Question:
$100$ independent measurements (i.e. samples) are made of a random variable, which has an exponential distribution $\lambda e^{−\lambda x}$, and their average is found. Consider the probability distribution of this average. The expectation value of this distribution will be $\lambda^{-1}$.
What is the standard deviation of the distribution?
End of Question.
Start of Solution:
The sum of the $100$ values of $x$ is $$X=\sum_{i=0}^{i=N}x_i$$ and has an expectation value $$E(x)=\sum_{i=0}^{i=N}E(x_i)=N\cdot E(x)$$
Since $E(x) \propto x$, the expectation value of the average $\overline{x}=\dfrac{X}{N}$ is $$E\Big(\overline{x}\Big)=\dfrac{N\cdot E(x)}{N}=E(x)=\color{blue}{\dfrac{1}{\lambda}}$$ $\color{blue}{\fbox{$\text{as stated in the question}$}}$. Similarly, for the Variance $$Var(x)=\sum_{i=1}^{i=N}Var(x_i)=N\cdot Var(x)$$ Since $Var(x)\propto x^2$, the variance of the average $\overline{x}=\dfrac{X}{N}$ is $$Var\Big(\overline{x}\Big)=\dfrac{N\cdot Var(x)}{N^2}=\dfrac{1}{N}\cdot \color{red}{Var(x)}=\dfrac{1}{100}\cdot \color{red}{\dfrac{1}{\lambda^2}}\tag{?}\label{}$$ The Standard Deviation is therefore $$\sqrt{Var\Big(\overline{x}\Big)}=\dfrac{1}{10\lambda}$$
End of Solution.
I understand why $E(x)=\color{blue}{\dfrac{1}{\lambda}}$ as it was stated in the question as shown by the $\color{blue}{\mathrm{blue}}$ box. But what I don't understand is why $\color{red}{Var(x)}\stackrel{\eqref{*}}=\color{red}{\dfrac{1}{\lambda^2}}$. Could someone please explain to me how this was worked out?
We have
$$E(X^2)=\int_0^\infty x^2 \lambda e^{-\lambda x}dx=\frac{2}{\lambda^2}$$
So,
$$Var(X)=E(X^2)-E(X)^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}$$