Question

Each phone call by Ali consumes an amount of time that follows an exponential distribution with mean 5 minutes. The number of different phone calls Ali makes on any given day has a Poisson distribution with mean 3. Assume that a single call always falls within a single day (no calls continue past midnight). Further, suppose that the number of phone calls that Ali makes on different days are independent, and that the lengths of the phone calls are also independent of each other. For simplicity, also assume that different phone calls never overlap and that there are 30 days in each given month. Let X be the total number of minutes Ali spends on the phone during one month. Find E(X) .

Answer 1

Answer:

Variance of X is 4500minutes

Step-by-step explanation:

Given that ,

 X = 30

$\sum_{i=1}^N X_i\\\\X_i \text {number of minutes in each call}\\\\n =\text {number of call in a given day}$

N≅P(3)

Xi ≅Exp(5)

$E(x)=E(30) \sum_{i=1}^N\\\\E(x)=30E(N).E(x)\\\\E(N)=3\\\\V(N)=3\\\E(x_1)=5\\\\ \text {Mean}= \text {variance of exponential}\\\\V_{x_1}=5^2=25\\\\V_{x1}=25\\\\E(x)=30*3*5\\\\E(x) = 450 minutes$

Now Variance is

⇒$Var{\sum_{i=1}^NX_i \\\\=E(N)Var(X_1)+E(x_1)^2Var(N)$

$V(x) = 30[\sum_{i=1}^NX_i]$

$=30[E(N)V(x_i)+E(x_1)^2V(N))\\\\=30[(3*25)+(25*3)]\\\\=30[75+75]\\\\=30*150\\\\V(x)=4500 \text {minutes}$