Answer:
The probability is [tex]P(X>30)=0.75[/tex]
Step-by-step explanation:
We know that the probability of testing the fire alarm during this period is uniformly distributed that is [tex]X[/tex] ~ [tex]U(0,120)[/tex].
We need to find [tex]P(X>30)[/tex]
Given a continuous random variable [tex]X[/tex] with distribution :
[tex]X[/tex] ~ [tex]U(a,b)[/tex] , where ''[tex]a[/tex]'' and ''[tex]b[/tex]'' are real numbers
The probability density function is :
[tex]f_{X}(x)=\frac{1}{b-a}[/tex] if x ∈ (a,b)
[tex]f_{X}(x)=0[/tex] if x ∉ (a,b)
In the exercise we have [tex]X[/tex] ~ [tex]U(0,120)[/tex] , therefore the probability density function is :
[tex]f_{X}(x)=\frac{1}{120}[/tex] if x ∈ (0,120)
[tex]f_{X}(x)=0[/tex] if x ∉ (0,120)
If we want to find [tex]P(X>30)[/tex] we need to perform the integral
[tex]\int\limits^i_d {f_{X}(x)} \, dx[/tex]
Where [tex]d=30[/tex] and ''[tex]i[/tex]'' represents + ∞
Now, given that [tex]f_{X}(x)[/tex] is 0 when x ∉ (0,120), we will need to integrate between 30 and 120 to find the probability.
If we perform this integral ⇒
[tex]\int\limits^e_d {\frac{1}{120}} \, dx[/tex]
Where [tex]d=30[/tex] and [tex]e=120[/tex] ⇒
[tex]\int\limits^e_d {\frac{1}{120}} \, dx=\frac{3}{4}=0.75[/tex]
