Expected value and variance of given probability density function

Question

Could someone please help me with these? I am a new member, so I had to hyperlink my pdf function. Please provide me with answer, I have an assasement test very soon.

A continuous random variable x has a probability density function:

\[ \mathrm{pdf}(x) = \frac{\exp\left(-\frac{(x-3)^2}{16}\right)}{\sqrt{16\pi}}, \qquad x\in (-\infty, \infty) \]

a) Find the expected value, variance and the standard deviation of continuous random variable.

b) Find the expected value of a random variable $\exp(-x/2)$

Answer 1

Continuing on Dilip's suggestion. PDF or a Normal RV is $$ f(x)=\frac{1}{\sqrt{2 \pi} \sigma}e^{-\big( \frac{x- \mu}{\sigma \sqrt{2}} \big)^2} $$ Using the values, you get $\mu=3, \sigma^2=8$, hence $X \sim N(3,8)$.

EDIT I probably misunderstood the OP's second question, as Dilip pointed out. Here's what I'd do. If $X \sim N(0,1)$, then: $$ \mathbf{E}e^{-\frac{x}{2}}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\frac{x}{2}}e^{-\frac{x^2}{2}}dx=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\big(\frac{x^2}{2}+\frac{x}{2})}dx=\frac{e^{\frac{1}{8}}}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}e^{-\frac{1}{2} \big(x+ \frac{1}{2} \big)^2}dx = e^{\frac{1}{8}} \cdot 1\\ $$ The last step is since the integral is $\mathbf{P}(-\infty < X<\infty)$=1, where $X \sim N(-\frac{1}{2},1)$