Question

b. What is the variance () 2. A particular probability function is given by f(x) - a. Normalize the function. b. What is the probability that x lies between 1 and 2? 0 elsewhere

I just need help with the first part.

Answer 1

we know that a positive function is a valid pdf if it integrates to 1 over the range

Let C be the normalizing constant for the given function f(x).

For f(x) to be a valid pdf we need

$\begin{align*} &\int_{-\infty}^{\infty}Cf(x)dx=1\\ \implies &\int_0^{\infty}Ce^{-3x}dx=1\\ \implies &C\left(\left.\frac{-e^{-3x}}{3}\right|_0^{\infty} \right )=1\\ \implies &C\left(0-\frac{-e^{-3\times 0}}{3} \right )=1\\ \implies &C\times \frac{1}{3}=1\\ \implies &C=3 \end{align*}$

a) The normalized function is

$\begin{align*} f(x)=\begin{cases}3e^{-3x},&x>0\\0,&\text{elsewhere} \end{cases} \end{align*}$

b) The probability that x lies between 1 and 2 is

$\begin{align*} P(1<X<2)&=\int_1^2f(x)dx\\ &=\int_1^23e^{-3x}dx\\ &=3\left(\left.\frac{-e^{-3x}}{3}\right|_1^2 \right )\\ &=\left(e^{-3\times 1}-e^{-3\times 2} \right )\\ &=0.0473 \end{align*}$

ans: The probability that x lies between 1 and 2 is 0.0473