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# Probability! watch

1. AA
2. If you let t* be the average length of time, then the average cost will just be:

Average length of call time x cost per minute.

F(t*) x cost per minute. Unless I'm missing something.
3. (Original post by little_wizard123)
If you let t* be the average length of time, then the average cost will just be:

Average length of call time x cost per minute.

F(t*) x cost per minute. Unless I'm missing something.
Thanks for your reply. But what would be the average cost? as in a number and not in terms of t*.
4. You can't solve it, unless you have some other information. Something like "you want to find the minimum average cost".
5. (Original post by little_wizard123)
You can't solve it, unless you have some other information. Something like "you want to find the minimum average cost".
I have the median time as 0.4142 minutes if thats of any use. And it just says find the average cost incurred from the typical caller. I'm pretty sure it wants a number for it but can't seem to figure out how to get it.
6. (Original post by JBKProductions)
The time that a customer is put on hold is given by F(t) = 1-(t+1)^-2 where t is measured in minutes. The company who runs the helpline incurs a cost of 6(t+1)^0.5 pence who is on hold for t minutes. Find the average cost.
I'm not sure how I can start this question off. can someone please give me a few hints please. Thanks.
F(t) is your distribution function right? Use the density function.
7. ^^ Makes more sense. I was assuming it was just a standard function.
8. (Original post by Dirac Delta Function)
F(t) is your distribution function right? Use the density function.
Thanks. I have the density function as 2(t+1)^-3. But I'm not sure what you would use it for. Do I set it to 0 and find the mode then sub that into the expression for cost? Or is that completely wrong?
EDIT: actually I realised it cant be solved....I'm completey stuck.
9. (Original post by JBKProductions)
Thanks. I have the density function as 2(t+1)^-3. But I'm not sure what you would use it for. Do I set it to 0 and find the mode then sub that into the expression? Or is that completely wrong?
so the density function f(t)=2(t+1)^-3 is a bit like the probability that it lasts t, (though of course it's not the probability, but it is useful to think of it like this informally).

If this were a discrete problem, and the probability of it lasting t were P(t), and the cost were c(t), how would you work out the expectation of the cost (if t were numbers 0,1,2,....)?
10. (Original post by Dirac Delta Function)
so the density function f(t)=2(t+1)^-3 is a bit like the probability that it lasts t, (though of course it's not the probability, but it is useful to think of it like this informally).

If this were a discrete problem, and the probability of it lasting t were P(t), and the cost were c(t), how would you work out the expectation of the cost (if t were numbers 0,1,2,....)?
Do I integrate from 0 to infinity P(t)*c(t)?
11. (Original post by JBKProductions)
Do I integrate from 0 to infinity P(t)*c(t)?
Well, for a discrete version, you would sum, but yes, in this case, you integrate

In general,

g(x) is a function of some random variable X, f(x) is the density function of X. In this case, X is time, and g(x) is c(x), the cost.
12. (Original post by Dirac Delta Function)
Well, for a discrete version, you would sum, but yes, in this case, you integrate

In general,

g(x) is a function of some random variable X, f(x) is the density function of X. In this case, X is time, and g(x) is c(x), the cost.
Thanks! Lol this is a bit random, but when I was searching about the density function I came across the Dirac Delta function so found it quite funny that you helped me with probability with that as you name. . Anyway thats the end of the randomness thanks again!
13. (Original post by JBKProductions)
Thanks! Lol this is a bit random, but when I was searching about the density function I came across the Dirac Delta function so found it quite funny that you helped me with probability with that as you name. . Anyway thats the end of the randomness thanks again!
No problem.

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