Answer:
a) 0.1353
b) 0.3679
Step-by-step explanation:
Let's start by defining the random variable T.
T : ''The time (in hours) required to repair a machine''
T ~ exp (λ)
T ~ exp (1)
The probability density function for the exponential distribution is
(In the equation I replaced λ = L)
[tex]f(x)=Le^{-Lx}[/tex]
With L > 0 and x ≥ 0
In this exercise λ = 1 ⇒
[tex]f(x)=e^{-x}[/tex]
For a)
[tex]P(T>2)[/tex]
[tex]P(T>2)=1-P(T\leq 2)[/tex]
[tex]P(T>2)=1-\int\limits^2_0 {e^{-x} } \, dx[/tex]
[tex]P(T>2)=1-(-e^{-2}+1)[/tex]
[tex]P(T>2)=e^{-2}=0.1353[/tex]
For b)
[tex]P(T\geq 10/T>9)[/tex]
The event (T ≥ 10 / T > 9) is equivalent to the event T ≥ 1 so they have the same probability of occur
[tex]P(T\geq 10/T>9)=P(T\geq 1)[/tex]
[tex]P(T\geq 1)=1-P(T<1)=1-\int\limits^1_0 {e^{-x} } \, dx[/tex]
[tex]P(T\geq 1)=1-(-e^{-1}+1)=e^{-1}=0.3679[/tex]
