Density function - probability

Question

My question is - If I have this density:

$$f_X(x) = \begin{cases} A, & 0 \le x \le 2 \\ B, & 2 < x \le 5 \\ 0, & \text{else} \end{cases}$$

and it is known that $E(x) = 3.05$ I need to find $P(0.05<X≤3|X>0.1)$

So how can I calculate A and B? Do I'm add them together to 3.05? and then do the formula of Bayes theorem?

Answer 1

take a look the informatio you have:

$$\begin{cases} 2A+3B=1 \\ A\int_0^2x dx+B\int_2^5x dx=3.05 \\ \end{cases} $$

$$\begin{cases} 2A+3B=1 \\ 2A+10.5B=3.05 \\ \end{cases} $$

that means:

$$\begin{cases} A=0.09 \\ B=0.273 \\ \end{cases} $$

EDIT: rest of the solution

First of all a drawing:

This is the graph of your density. As you can check, the total area is correctly 1 (but some decimals)

Your probability states

$$\mathbb{P}[0.05 <X <3|X>0.1]=\frac{\mathbb{P}[0.1 <X <3]}{\mathbb{P}[X>0.1]}=\frac{(2-0.1)\cdot 0.09+(3-2)\cdot 0.273}{(2-0.1)\cdot 0.09+(5-2)\cdot 0.273}\approx 0.448$$

Answer 2

I am assuming that your $f_X$ is a probability density function (PDF).

Note that the integral (over the possible range of the data $X$) of any PDF is 1 (since probabilities sum to 1). This is your first condition.

Also, you have a value for the expectation. This is your second condition.

Now, you can find A and B (two unknowns) using the two conditions above.

To find the conditional probability you stated, use Bayes Theorem.