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After measuring the duration of many telephone calls, the telephone company found their data was well-approximated by the density function p(x) = 0.4e^(−0.4x), where x is the duration of a call, in minutes.
a) What percentage of calls last between 4 and 5 minutes? (nearest tenth)
b) What percentage of calls last 4 minutes or less? (nearest tenth)
c) What percentage of calls last 4 minutes or more? (nearest tenth)

Respuesta :

Answer:

a) [tex] P(4<X<5)= \int_{4}^5 0.4 e^{-0.4 x} dx = 0.4 \int_{4}^5 e^{-0.4 x} dx[/tex]

[tex] P(4<X<5)= -e^{-0.4x} \Big|_4^5 = -e^{-0.4*5} + e^{-0.4*4} = -0.135+0.202=0.07 \approx 0.1[/tex]

b) [tex] P(0<X<4)= \int_{0}^4 0.4 e^{-0.4 x} dx = 0.4 \int_{0}^4 e^{-0.4 x} dx[/tex]

[tex] P(0<X<4)= -e^{-0.4x} \Big|_0^4 = -e^{-0.4*4} + e^{-0.4*0} = -0.202+1=0.798 \approx 0.8[/tex]

c) [tex] P(X>4)= 1-P(X<4) = 1-P(0<X<4)=1-0.798 = 0.202\approx 0.2[/tex]

Step-by-step explanation:

For this case we define the random variable X who represent the duration of a call, in minutes, and the density function for X is given by:

[tex] f(x) = 0.4 e^{-0.4 x}[/tex]

Part a

For this case we want to find this probability:

[tex] P(4<X<5)[/tex]

And we can find this probability with this integral:

[tex] P(4<X<5)= \int_{4}^5 0.4 e^{-0.4 x} dx = 0.4 \int_{4}^5 e^{-0.4 x} dx[/tex]

[tex] P(4<X<5)= -e^{-0.4x} \Big|_4^5 = -e^{-0.4*5} + e^{-0.4*4} = -0.135+0.202=0.07 \approx 0.1[/tex]

Part b

[tex] P(0<X<4)= \int_{0}^4 0.4 e^{-0.4 x} dx = 0.4 \int_{0}^4 e^{-0.4 x} dx[/tex]

[tex] P(0<X<4)= -e^{-0.4x} \Big|_0^4 = -e^{-0.4*4} + e^{-0.4*0} = -0.202+1=0.798 \approx 0.8[/tex]

Part c

For this case we want this probability:

[tex] P(X>4)[/tex]

And we can use the complement rule and the result from part b and we got:

[tex] P(X>4)= 1-P(X<4) = 1-P(0<X<4)=1-0.798 = 0.202\approx 0.2[/tex]

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