# Homework Help: Probability - I am so lost here

1. Apr 22, 2007

Question:
Long-distance calling plan A offers a flat rate service at 10 cents per minute. Calling plan B charges 99 cents for every call under 20 minutes; for calls over 20 minutes, the charge is 99 cents for the first 20 minutes plus 10 cents for every additional minute. (Note that these plans measure your call duration exactly, without rounding to the next minute or even second.) If your long-distance calls have exponential distribution with expected value $\tau$ minutes, which plan offers a lower expected cost per call?

I am lost. Here is my attempt.

We first want a random variable that yields the cost for each plan. Lets first let $T$ be a random variable that represent the time of the call. Thus,

$$C_A = 10 T$$
A random variable of cost is representative of the random variable for the cost of the time.

Now, I don't really know how to do plan B, here is my attempt:
$$C_B = 99 + 10(T-20)u(T-20)$$ where $u(T-20)$ would be the step function. I am trying to say that the $10(T-20)$ term would turn on at $T \geq 20$.

Now I want the expected value of each, thus:

$$E(C_A) = 10 \tau$$
$$E(C_B) = 99 + 10E((T-20)u(T-20))$$

I am lost, and need help. Thanks!

2. Apr 22, 2007

### ZioX

What you want is:

$$E[C_B]=\int_0^{20}.99f_X(x|\tau=t)dx+\int_{20}^{+\infty}(.10(x-20)+.99)f_X(x|\tau=t)dx$$

Now, you compare values of tau to see what one will be the best plan.

Edit: This is the right one. ;0

Note here how if you take that .99f(x) integrated from 20 to infinity out and combine it with the first summand you get your original equation which is actually more easier to compute.

So, essentially it just ammounts to finding E[(t-20)u(t-20)] which is precisely $$\int_{20}^{+\infty}(t-20)f_X(t)dt$$.

Last edited: Apr 22, 2007
3. Apr 22, 2007

Last edited: Apr 22, 2007
4. Apr 22, 2007

### ZioX

Sorry, X is the r.v. representing time on the phone. $$f_X(x|\tau=t)$$ denotes the distribution of x given tau=t. I changed it to that because I wanted to emphasize that expected cost was a function of tau.

Last edited: Apr 22, 2007
5. Apr 22, 2007

No worries!

I still have a few questions though. So, how does

$$E[C_B]=\int_0^{20}.99f_X(x|\tau=t)dx+\int_{20}^{+\infty} .99(x-20)f_X(x|\tau=t)dx$$

account for the 10 cents per minute charge after 20 minutes? Is it supposed to read:

$$E[C_B]=\int_0^{20}.99f_X(x|\tau=t)dx+\int_{20}^{+\infty} .10(x-20)f_X(x|\tau=t)dx$$

?

If this is the case, how do you get
$$E[C_B]=\int_0^{20}.99f_X(x|\tau=t)dx$$? I don't understand that term.

Thanks!

6. Apr 22, 2007

### ZioX

I'm sorry, I made a mistake. I fixed it up above.

7. Apr 22, 2007

No worries.

So where does,
$$\int_0^{20}.99f_X(x|\tau=t)dx$$

come from?

8. Apr 22, 2007

### ZioX

Estimated cost is really just integrating what you're paying at time t multiplied by the pdf evaluated at t.

So that integral comes from the fact that from 0 to 20 minutes we're paying a constant 99 cents. After 20 minutes we are paying 10 cents for every minute after 20 minutes plus the additional 99 cents we already used for the first 20 minutes. That is where (.10(x-20)+.99)f(x) comes from.

9. Apr 22, 2007