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    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.
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    (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*.
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    You can't solve it, unless you have some other information. Something like "you want to find the minimum average cost".
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    (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.
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    (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.
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    ^^ Makes more sense. I was assuming it was just a standard function.
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    (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.
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    (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,....)?
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    (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)?
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    (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

    \int_{0}^{\infty}c(t)f(t)dt

    In general, E[g(X)] = \int_{-\infty}^{\infty}g(x)f(x)dx

    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.
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    (Original post by Dirac Delta Function)
    Well, for a discrete version, you would sum, but yes, in this case, you integrate

    \int_{0}^{\infty}c(t)f(t)dt

    In general, E[g(X)] = \int_{-\infty}^{\infty}g(x)f(x)dx

    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!
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    (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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