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

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

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)=(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)=(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 {(1)/(120)} \, dx`

Where
`d=30` and
`e=120` ⇒

`\int\limits^e_d {(1)/(120)} \, dx=(3)/(4)=0.75`