Solution Found!
The duration Y of long-distance telephone calls (in
Chapter 4, Problem 156E(choose chapter or problem)
The duration 𝑌 of long-distance telephone calls (in minutes) monitored by a station is a random
variable with the properties that
\(P(Y=3)=.2\quad\ \text{ and }\quad\ P(Y=6)=.1.\)
Otherwise, 𝑌 has a continuous density function given by
\(f(y)=\left\{\begin{array}{ll}
(1 / 4) y e^{-y / 2}, & y>0, \\
0, & \text { elsewhere. }
\end{array}\right.
\)
The discrete points at 3 and 6 are due to the fact that the length of the call is announced to the
caller in three-minute intervals and the caller must pay for three minutes even if he talks less
than three minutes. Find the expected duration of a randomly selected long-distance call.
Equation Transcription:
Text Transcription:
P(Y=3)=.2 and P(Y=6)=.1.
f(y)=
(1/4)ye-y/2, y>0,
0, elsewhere.
Questions & Answers
QUESTION:
The duration 𝑌 of long-distance telephone calls (in minutes) monitored by a station is a random
variable with the properties that
\(P(Y=3)=.2\quad\ \text{ and }\quad\ P(Y=6)=.1.\)
Otherwise, 𝑌 has a continuous density function given by
\(f(y)=\left\{\begin{array}{ll}
(1 / 4) y e^{-y / 2}, & y>0, \\
0, & \text { elsewhere. }
\end{array}\right.
\)
The discrete points at 3 and 6 are due to the fact that the length of the call is announced to the
caller in three-minute intervals and the caller must pay for three minutes even if he talks less
than three minutes. Find the expected duration of a randomly selected long-distance call.
Equation Transcription:
Text Transcription:
P(Y=3)=.2 and P(Y=6)=.1.
f(y)=
(1/4)ye-y/2, y>0,
0, elsewhere.
ANSWER:
Solution:
Step 1 of 2:
It is given that Y is a random variable denoting the duration of long distance calls. For discrete Y,
P(Y=3)=0.2 and P(Y=6)=0.1.
For continuous Y, the density function is given as
f(y)=
Using this, we need to find the expected duration of a randomly selected long distance call.