Respuesta :
The question is incomplete. Here is the complete question.
Nite Time Inn has a toll-free telephone number so that customers can call at any time to make a reservation. A typical call takes about 4 minutes to complete, and the time required follows an exponential distribution. find the probability that a call takes
a) 3 minutes or less
b) 4 minutes of less
c) 5 minutes of less
d) Longer than 5 minutes
e) Longer than 7 minutes
Answer: a) P(X<3) = 0.882
b) P(X<4) = 0.908
c) P(X<5) = 0.928
d) P(X>5) = 0.286
e) P(X>7) = 0.174
Step-by-step explanation: Exponential distribution is related with teh amount of time until some specific event happens.
If X is a continuous random variable, probability is calculated as:
[tex]P(X<x) = 1-me^{-mx}[/tex]
in which:
m is decay parameter, given by: [tex]m=\frac{1}{mean}[/tex]
For the Nite Time Inn calls:
[tex]m=\frac{1}{4}[/tex]
m = 0.25
(a) P(X<3)
[tex]P(X<3) = 1-0.25e^{-0.25.3}[/tex]
[tex]P(X<3) = 1-0.25e^{-0.75}[/tex]
[tex]P(X<3) = 1-0.25*0.472[/tex]
P(X < 3) = 0.882
The probability the call takes less than 3 minutes is 0.882.
(b) P(X<4)
[tex]P(X<4) = 1-0.25e^{-0.25.4}[/tex]
[tex]P(X<4) = 1-0.25e^{-1}[/tex]
P(X < 4) = 0.908
The probability the call takes less tahn 4 minutes is 0.908.
(c) P(X<5)
[tex]P(X<5) = 1-0.25e^{-0.25.5}[/tex]
[tex]P(X<5) = 1-0.25e^{-1.25}[/tex]
P(X < 5) = 0.928
The probability of calls taking less than 5 minutes is 0.928.
(d) P(X>5)
Knowing that the sum of probabilities of less than and more than has to equal 1:
P(X<x) + P(X>x) = 1
P(X>x) = 1 - P(PX<x)
[tex]P(X>x) = 1-(1-me^{-m*x})[/tex]
[tex]P(X>x)=me^{-mx}[/tex]
For P(X>5):
[tex]P(X>5) = 0.25e^{-1.25}[/tex]
P(X > 5) = 0.286
The probability of calls taking more than 5 minutes is 0.286.
(e) P(X>7)
[tex]P(X>7)=0.25e^{-0.25.7}[/tex]
[tex]P(X>7)=0.25e^{-1.75}[/tex]
P(X > 7) = 0.174
The probability of calls taking more than 7 minutes is 0.174.
