IX is uniformly distributed over the interval [0,10], compute P{2

(i)

$P(2<X<9)=\displaystyle\int_{2}^{9}\dfrac{1}{10-0}dx=[0.1x]\begin{matrix} 9 \\ 2 \end{matrix}=0.7$

(ii)

$P(1<X<4)=\displaystyle\int_{1}^{4}\dfrac{1}{10-0}dx=[0.1x]\begin{matrix} 4 \\ 1 \end{matrix}=0.3$

(iii)

$P(X<5)=\displaystyle\int_{0}^{5}\dfrac{1}{10-0}dx=[0.1x]\begin{matrix} 5 \\ 0 \end{matrix}=0.5$