Calculating interval probability when both values are above/below upper/lower bounds

Question

I am bit rusty on probability and I was refreshing on uniform probability distribution.

I wanted to calculate $P( 0.5 < x \leq 1.5 )$ given Uniform$(0, 4)$.

I would normally follow the formula, that would give $\frac{ 4 - 0 }{ 1.5 - 0.5 } $

But it doesn't seem right to have probability of $4$.

Should the upper bound be trimmed? How would you approach this problem?

Answer 1

Uniform $(0;4)$ has a rectangular density wiht base 4 and height $\frac{1}{4}$ so the requested probability is

$$(1.5-.5)\times\frac{1}{4}=\frac{1}{4}$$

This is a useful drawing

Answer 2

The cumulative distribution function of the uniform distribution $U(a,b)$ is:

$$F(x)=\begin{cases} 0 \space\space\space\space\text{for $x<a$} \space\space \\\frac{x-a}{b-a} \space\space\space\space\text{for $a\leq x \leq b$} \space\space\ \\1 \space\space\space\space\text{for $x>b$} \space\space \end{cases} $$

In your case $a=0$ and $b=4$ so what is $F(1.5)-F(0.5)$?

Alternatively you have that $P(0.5<x \leq 1.5)=\int_{0.5}^{1.5}f(x)dx=\int_{0.5}^{1.5}\frac{1}{4}dx.$