Given the density function: $\frac{1}{2}\exp\left(-\frac{x}{2}\right), \space x > 0$ find $P\left(\sum_{i=1}^{81}X_i > 170\right)$

Question

Suppose that $X_1,X_2...X_{81}$ are independent random variable with the same probability density function $$\frac{1}{2}\exp\left(-\frac{x}{2}\right), \space x > 0$$

Find $$P\left(\sum_{i=1}^{81}X_i > 170\right)$$

My problem is figuring out the expected value and variance. Is it just the coefficient of what's inside the exp brackets? Not too sure about this one.

Answer 1

We have $E(X_1+\cdots +X_n)=E(X_1)+\cdots +E(X_n)$, and, by independence, $\text{Var}(X_1+\cdots+X_n)=\text{Var}(X_1)+\cdots +\text{Var}(X_n)$.

The $X_i$ have exponential distribution parameter $\lambda=\frac{1}{2}$. It is a standard fact that the mean of such an exponentially distributed random variable is $\frac{1}{\lambda}$, and the variance is $\frac{1}{\lambda^2}$.

Remark: If proof for the "standard facts" is needed, for the mean calculate $\int_0^\infty \lambda xe^{-\lambda x}\,dx$, using integration by parts. For the variance of such a random variable $X$, use the fact that the variance is $E(X^2)-(E(X))^2$, and find $E(X^2)$ by integration by parts.