1 minute of call costs 1 currency. Let's assume that duration of call has exponential distribution with a parameter $\lambda=1$, which implies that call lasts for an avarage of 1 minute.
How much on avarage do we pay for a call?
$X$-Duration of call
$Z$-Costs of call
My take on it:
$$ \lambda=1\Rightarrow \left\{\begin{matrix} P(X=t)=e^{-t}\\ P(X\leq t)=1-e^{-t} \end{matrix}\right.$$ $$ for$$ $$t\leq0$$
$$P(X=t)=\lambda e^{-\lambda t}=e^{-t},t>0$$
For call we pay $z$ currency $(z=1,2,...)$ if call lasts $z-1<X\leq z$
$$P(Z=z)=P(z-1<X\leq z)=(1-e^{-z})-(1-e^{-(z-1)})=e^{1-z}-e^{z}$$
Avarage costs of call $E(Z)$
$E(Z)=\sum_{z=1}^{\infty}z\times P(Z=z)=\sum_{z=1}^{\infty}z(e^{1-z}-e^{-z})=(e-1)\sum_{z=1}^{\infty}ze^{-z}=(e-1)\times \frac{e}{(e-1)^{2}}$
$\approx1.68$ currency
Could anyone check if it makes sense?