Answer:
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 = \(t / \mu_F\)
=> \(E(n) = t / \mu_F\)