Question

Your school wants to test the fire alarm system during the two hour period between 9:00 a.M. And 11:00 a.M. The probability of testing the fire alarm during this period is uniformly distributed that is X∼U(0,120). Find the probability that you have to wait more than 30 minutes, that is, find P(X>30).

Answer 1

Answer:

The probability is $P(X>30)=0.75$

Step-by-step explanation:

We know that the probability of testing the fire alarm during this period is uniformly distributed that is $X$ ~ $U(0,120)$.

We need to find $P(X>30)$

Given a continuous random variable $X$ with distribution :

$X$ ~ $U(a,b)$ , where ''$a$'' and ''$b$'' are real numbers

The probability density function is :

$f_{X}(x)=\frac{1}{b-a}$    if x ∈ (a,b)

$f_{X}(x)=0$  if x ∉ (a,b)

In the exercise we have $X$ ~ $U(0,120)$ , therefore the probability density function is :

$f_{X}(x)=\frac{1}{120}$ if x ∈ (0,120)

$f_{X}(x)=0$ if x ∉ (0,120)

If we want to find $P(X>30)$ we need to perform the integral

$\int\limits^i_d {f_{X}(x)} \, dx$

Where $d=30$ and ''$i$'' represents + ∞

Now, given that $f_{X}(x)$ is 0 when x ∉ (0,120), we will need to integrate between 30 and 120 to find the probability.

If we perform this integral ⇒

$\int\limits^e_d {\frac{1}{120}} \, dx$

Where $d=30$ and $e=120$ ⇒

$\int\limits^e_d {\frac{1}{120}} \, dx=\frac{3}{4}=0.75$