1

The time one has to wait $T$ (in hours) between the telephone calls one recieves is random with function $f(t) = (3/2)e^{-\frac{3}{2}t}$ for $t>=0$.

  1. Find the expected time between recieved calls and find the corresponding variance.
  2. For two independent waiting-times, $T_1$ and $T_2$, find the probability that both are greater than 1.

(1)

We can see that $T$ has an exponential distiribution with parameter $\lambda = \frac{3}{2}$ for a $t >= 0$. The expected time and variace are given by:

\begin{align}\\ & E(X)=\mu _X=\alpha\beta=\frac{1}{\lambda}=\frac{1}{\frac{3}{2}}=\frac{2}{3} \\ & V(X)=\sigma ^2_X=\alpha\beta^2=\frac{1}{(\frac{3}{2})^2}=\frac{4}{9} \\ \end{align}

2

We have:

\begin{align}\\ & \lambda=\frac{3}{2} && F(x;\lambda)=1-e^{-\lambda x} && \text{for} x\geqslant0 \\ \end{align}

and:

\begin{align}\\ & P(X > x) = 1 - P(X\leqslant x) \\ & P(X > x) = 1 - F(x;\lambda ) \\ & P(X > x) = 1 - (1-e^{-\lambda x}) \\ & P(X > x) = e^{-\lambda x} \end{align}

But I don't know how to continue if I even have understood it correctly so far..

Mampenda
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1 Answers1

1

Part (1) is correct. For part (2), all you have to do is compute the probability that $T_1 > 1$, i.e. $$\Pr[T_1 > t] = e^{-3t/2},$$ and then observe that since the times are independent and identically distributed, the probability that both $T_1$ and $T_2$ are greater than $1$ is simply the square of this probability.

heropup
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