The standard deviation for a random variable with probability density function $f$ and mean $\mu$ is defined
$$\sigma = \left[ \int _ { - \infty } ^ { \infty } ( x - \mu ) ^ { 2 } f ( x ) d x \right] ^ { 1 / 2 }$$
Find the standard deviation for an exponential density function with mean $10 .$
A) $3.3$
B) 2
C) 5
D) $2.5$
E) 10

Question

Quizplus · Accepted Answer

For an exponential density function with parameter $\lambda$, the mean is given by $\mu = \frac{1}{\lambda}$. Therefore, for a mean of $10$, we have $\frac{1}{\lambda} = 10$ or $\lambda = \frac{1}{10}$.
The density function is given by $f(x) = \lambda e^{-\lambda x}$ for $x\geq 0$ and $f(x) = 0$ for $x<0$.
Using the formula for standard deviation, we have
\begin{align*}
\sigma &= \sqrt{\int_{-\infty}^{\infty}(x-\mu)^2f(x)dx}\
&= \sqrt{\int_{0}^{\infty}(x-10)^2\left(\frac{1}{10}e^{-\frac{1}{10}x}ight)dx}\
&= \sqrt{\frac{1}{10}\int_{0}^{\infty}(x-10)^2e^{-\frac{1}{10}x}dx}\
&= \sqrt{\frac{1}{10}\left(\frac{2}{100}ight)} \quad 	ext{(using integration by parts)}\
&= \sqrt{\frac{1}{500}} \
&= \boxed{	extbf{(E) }10^{-1/2}}
\end{align*}

The Standard Deviation for a Random Variable with Probability Density fff

The Standard Deviation for a Random Variable with Probability Density $f$