What is the amount that the company expects to pay for the 72 customers on the airplane altogether? (Of course their waiting times are all the same.)

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Suppose that, when an airplane waits on the runway, the company must pay each customer a fee if the waiting time exceeds $3$ hours. Suppose that an airplane with $72$ passengers waits an exponential amount of time on the runway, with average $1.5$ hours. If the waiting time X, in hours, is bigger than $3$, then the company pays each customer $(100)(X − 3)$ dollars (otherwise, the company pays nothing). What is the amount that the company expects to pay for the $72$ customers on the airplane altogether? (Of course their waiting times are all the same.)

so $\lambda = 2/3$

and $P(X>3) = e^{-2} \approx 0.1353$. this is the probability that any individual passenger waits longer then 3 hours I.E. the probability that the airport has to pay them.

But how do I use this probability to calculate the expected amount that the company pays for the 72 customers? Especially because we don't know what specific time over $3$ hours an customer waits on average?

The final answer in the book says $1461.62

Can someone tell me the methodology to get this answer? What am I missing?

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The crux of the question is to recognise that the cost to the airline is given by $$N\int_3^{\infty} C(x)f(x)dx$$ where $$C(x) = 100(x-3)$$ is the penalty paid out to each customer in dollars, $$ f(x) = \frac{2}{3}\exp\left(-\frac{2x}{3}\right)$$ is the exponential probability distribution (with the rate $\lambda$ identified as $2/3$), and $N$ is the number of passengers, $72$.

Essentially, we're weighting the probability distribution by the cost function i.e. we're calculating the expectation of that cost using the probability distribution. The fact there are 72 passengers is uninteresting; we're calculating the expected payout per passenger and just scaling up.

Therefore, compute $$ 72\int_3^{\infty} 100(x-3)\frac{2}{3}\exp\left(-\frac{2x}{3}\right)dx$$ or, equivalently, $$ 4800\int_3^{\infty}(x-3)\exp\left(-\frac{2x}{3}\right)dx$$ using the facts that $$\int \exp(ax)dx = \frac{1}{a}\exp(ax)$$ and $$ \int x\exp(ax)dx = \frac{1}{a}\exp(ax) - \frac{1}{a^2}\exp(ax)$$

Note: Be wary when inputting $\infty$ in the limits; instead, consider what happens as $x \to \infty$.

This will give you the desired result of $1461.62.