The times between the arrivals of customers at a taxi stand are independent and have a distribution F with mean F. Assume an unlimited supply of cabs, such as might occur at an airport. Suppose that each customer pays a random fare with distribution G and mean G. Let W. t/ be the total fares paid up to time t. Find limt!1EW. t/=t.

Check the explanation

Step-by-step explanation:

Let

  \(W(t) = W_1 + W_2 + ... + W_n\)  

where W_i denotes the individual fare of the customer.

All W_i are independent of each other.

By formula for random sums,

E(W(t)) = E(Wi) * E(n)

  \(E(Wi) = \mu_G\)  

Mean inter arrival time =    \(\mu_F\)  

Therefore, mean number of customers per unit time =    \(1 / \mu_F\)  

=> mean number of customers in t time =    \(t / \mu_F\)  

=>    \(E(n) = t / \mu_F\)  

the correct answer is 828.07