I am reading Probability and Statistics for Engineering and the Sciences.
Exercise 63, Chapter 4 says:
A consumer is trying to decide between two long-distance calling plans. The first one charges a flat rate of 10¢ per minute, whereas the second charges a flat rate of 99¢ for calls up to 20 minutes in duration and then 10¢ for each additional minute exceeding 20 (assume that calls lasting a noninteger number of minutes are charged proportionately to a whole-minute’s charge). Suppose the consumer’s distribution of call duration is exponential with parameter $\lambda$.
- Which plan is better if expected call duration is 10 minutes? 15 minutes?
Assuming the first question, when the duration is 10 minutes, I computed the cost of the first plan as:
$h_1(x) = 10 * E[x] = 10 * 10 = 100$
However, how do I compute the cost for the second plan ($h_2(x)$) ?
I tried with:
$h_2(x) = 99 * F(x \leq 20) + 10 * (1 – F(x \leq 20)) \approx 87$
But the correct result is $112.53$.
Best Answer
In the second plan, you pay $99$ cents for sure for a call, plus a charge for calls that last more than $20$ minutes. Let us compute the expected value of this additional charge.
The additional charge occurs with probability $e^{-20/10}$.
Given that a call lasted more than $20$ minutes, the expected additional time is $10$, by the memorylessness property of the exponential distribution. Thus the expected additional charge is $(10)(e^{-20/10})(10)$.
So the expected cost of a call on the second plan, when mean length is $10$, is $99+(10)(e^{-20/10})(10)$ cents.
The calculation when the mean is $15$ is done in the same way.