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.

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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 =    \(t / \mu_F\)  

=>    \(E(n) = t / \mu_F\)  

2^y = ln 5, or   y(ln 2) = ln 5.   solving for y,   y = (ln 5) / (ln 2).   y=5

step-by-step explanation:

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