The diameter of sand grains in a special sand roof, measured in mm, can be considered as a random variable X with probability density $$f(x)=\begin{cases} k(x-x^2) & \text{if 0≤x≤1 } \\ \newline 0 &\text{otherwhise} \end{cases}$$
A) Determine the constant k so that this becomes a valid probability density
$1=k\int _0^1x-x^2dx$
$=\int _0^1xdx-\int _0^1x^2dx$
$\int _0^1xdx = \frac{1}{2} $
$\int _0^1x^2dx = \frac{1}{3}$
$=\frac{1}{2}-\frac{1}{3}= \frac{1}{6}$
$k = 0.16666$
B)Find the expectation $E (4X + 3)$ and the variance $V (4X + 3)$
Does anyone know how to go about these two questions? on a) I am pretty sure that I got the correct answer. on b I'm pretty much clueless...
any tips/solutions? thanks in advance
No. You dropped $k$ from the equation between step 1 an 2. You need to keep it in the expression, so... $$1=k(\tfrac 12-\tfrac 13)$$
Therefore $k=6$.
Use the rules of Linearity and Bilinearity.
$$\mathsf E(4X+3) = 4\mathsf E(X)+3\\\mathsf{Var}(4X+3)=4^2\mathsf{Var}(X)$$
Then you just need to use the definitions and integration to find the expectation and variance for $X$.