Question

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)

Answer 1

a)
`P(4<X<5)= \int_(4)^5 0.4 e^(-0.4 x) dx = 0.4 \int_(4)^5 e^(-0.4 x) dx`

`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`

b)
`P(0<X<4)= \int_(0)^4 0.4 e^(-0.4 x) dx = 0.4 \int_(0)^4 e^(-0.4 x) dx`

`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`

c)
`P(X>4)= 1-P(X<4) = 1-P(0<X<4)=1-0.798 = 0.202\approx 0.2`

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:

`f(x) = 0.4 e^(-0.4 x)`

Part a

For this case we want to find this probability:

`P(4<X<5)`

And we can find this probability with this integral:

`P(4<X<5)= \int_(4)^5 0.4 e^(-0.4 x) dx = 0.4 \int_(4)^5 e^(-0.4 x) dx`

`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`

Part b

`P(0<X<4)= \int_(0)^4 0.4 e^(-0.4 x) dx = 0.4 \int_(0)^4 e^(-0.4 x) dx`

`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`

Part c

For this case we want this probability:

`P(X>4)`

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

`P(X>4)= 1-P(X<4) = 1-P(0<X<4)=1-0.798 = 0.202\approx 0.2`