Question

$f\left(x\right)$,a continuous probability function, is equal to $\frac{1}{12}$and the function is restricted to $0\le x\le 12$. What is $P(0<x<12)$?

Short answer

The$P(0<x<12)$is$1$

Step-by-step solution

Given Information 

Given in the question that, the $f\left(x\right)$remains a continuous probability function. It is equal to $\frac{1}{12}$and the function is restricted to $0\le x\le 12$.

We have to find what is $P(6\le x\le 8)$.

Concept Introduction 

Consider the data in the question; the continuous probability function is written as follows:

$f\left(x\right)=\left\{\begin{array}{l}\frac{1}{12},\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}0\le x\le 12\\ 0,\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{otherwise}\end{array}\right.$

Here, $X$follows a uniform distribution.

Therefore, the area between $(0<x<12)$creates a rectangle with $x$axis.

Then the base is $\text{base}=(12-0)$and height is $\text{height}=\frac{1}{12}$

Calculation

For continuous random variable X, $P(c<x<d)=P(c\le x\le d)$.

The value of $P(0<x<12)$is:

$P(0<x<12)=\text{base}\times \text{height}$

$=(12-0)\times \frac{1}{12}$

$=\frac{12}{12}$

$=1$