find $\varepsilon$ such that $P(X\geq a+\varepsilon \land X\leq b - \varepsilon)=0.95$, with $X$ is uniformly distributed.

Question

Sketching the situation using the density function of the uniform distribution on $[a,b]$ its clear to me that is just a $0.025$ of $V:=∫_Ω \mathbb 1_Ω(x)\ dx$ it can't be simply $\varepsilon:= a+b\cdot 0.025$ or can it? Since I'm unsure if this holds when $\Omega \subset \mathbb R^n$. Is the skech even correct?

Answer 1

The given probability is just $\frac {(b-\epsilon) -(a+\epsilon)} {b-a}$. Setting this equal to $0.95$ and solving for $\epsilon$ gives $\epsilon=(0.025) (b-a)$.