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    I still can't get this - can anyone help ?

    A car rental company receives cars at n = 1000 cars/year for the first 5 years and none thereafter.

    The pdf for retiring a car is,

    Code:
    p(t) = (1/5)(1 - exp(-t/3))             , 0<t<5
         = (1/5)(exp[-(t-5)/3] - exp(-t/3)) , t > 5
    Show that after 6 years cars are being retired from the scheme at approxiamtely 11 per week.

    To answer that I did,

    N = int[t1 to t2] t*p(t) dt

    where
    t1 = 6 (years)
    t2 = 6 1/52 (years)
    and
    N would be the number of retired cars during that week.

    I thought that was how I was supposed to do it. Is that correct ?

    When I worked out that integral, I got N = 0.0134.

    How do I get 11 as the answer ?
    Where does the n = 1000 come into it, if at all ?
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    OK, it seems that method isn't right. It should be done like this.

    n = N*p(T)

    where,
    T is the time (yrs) after the start of the program
    N is the total number of cars supplied up to the time T (here N = 5000)
    n is the number of cars retired at time t = T (cars per year)
    p(t) is the pdf for retiring a car after t years from the start of the program
    T = 6
    N = 5000 (1000 cars/yr for 5 years)

    p(6) = 0.11624
    n = 5000*0.11624
    n = 581.2 cars/yr
    n = 11.18 cars/week
    n = 11 cars/week (approx)
    =============
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    Hi

    I helped you yesterday with a very similar stats question and we were both pretty sure the method was right, my method was probably wrong having not done stats for ages but I cant see exactly why it is wrong and I can't see how the above method is right, although it gives the right answer. Can you explain it to me just out of interest?

    Many thanks
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    (Original post by Kernel Rev)
    Hi

    I helped you yesterday with a very similar stats question and we were both pretty sure the method was right, my method was probably wrong having not done stats for ages but I cant see exactly why it is wrong and I can't see how the above method is right, although it gives the right answer. Can you explain it to me just out of interest?

    Many thanks
    Yes. my other post and this one are both for the same problem. My first post was just a paraphrase of the orginal problem as I was just interested in that one part of it and didn't think I would need to write it all out.

    I really do wish I could explain how that method works, but I basically suck at stats . I have no idea. I'm fairly/very good at other maths stuff, but not stats. I got that solution/method from another web-site.
    Perhaps if you read the posts there, it may explain things. I still don't (fully) understand the things he said!!

    The full question is given in post #4 on that site.

    Uhh, you wouldn't have an idea about part (b) would you??
    I'd be more than happy to help you with any maths queries.

    If I could get the pdf for part (b), I'm pretty sure I could do the convolution required for part (c).

    If it helps, the mark scheme for that question is,

    (a) [5]
    (b) [4]
    (c) [10]
    (d) [6]
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    Yep I feel exactly the same way as you do! I can do very well on statistics modules but thats not with out brute force and hard work whereas other maths disciplines I can sail through. The logic of Stats just baffles me but theres always a good feeling of satisfaction when you get a problem out!

    This question is still baffelling me though, I don't quite understand what the guy said on the other forum but I have a better understanding of the question now I think so I shall attempt what I can!

    (a) What is the mean value <t> of the time from delivery of a car until it is retired?

    By mean time we mean the expected time that a car will retire, by definition of expectation this is just the integral over all t, [0,inifinity] of t*p(t) which I work out to be 2years

    (b) Assuming that the cars are supplied at a constant rate, what is the probability density ?sup(t) that a given car will be supplied t years after the start of the program ?

    Now, we let X be the random variable "times after start of program that a car is supplied". We also know that the cars are supplied at a constant rate, one every 1/1000th of a year. At time t, 1000*t cars have been delivered and for every time up and til time t it was equiprobable for each of the cars deivered up to that point to have been delivered at that time so the probability of a car being delivered at time t is 1/1000*t also the p(T less than or equal t)=1 where T is the random variable "times after start of program that a car is supplied". The p.d.f which is what you want is not the probability though. This is about as far as I can get but if you have any more suggestions im gonna keep trying for a bit!
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    (Original post by Kernel Rev)
    Yep I feel exactly the same way as you do! ...!
    Well, at least I'm not alone in feeling that way

    As for part (a) I get <t> = τ = 3 years

    &lt;t&gt; = \int_0^\infty\ te^{-t/\tau}/\tau\ dt\ =\  \frac{1}{\tau}\int_0^\infty\ te^{-t/\tau}\ dt
    using u=t/\tau then t = u\tau and \tau\ du = dt
    substituting for t and dt,
    &lt;t&gt; = \frac{1}{\tau}\int_0^\infty\ u\tau e^{-u}\ \tau\ du
    &lt;t&gt; = \tau\int_0^\infty ue^{-u}\ du
    &lt;t&gt; = \tau\[-ue^{-u}\]_0^\infty+ \tau\int_0^\infty e^{-u}\ du
    &lt;t&gt; = \tau\{0+0\} + \tau\[-e^{-u}\]_0^\infty = \tau\{0+1\}
    &lt;t&gt; = \tau = 3
    =========

    I took the integral from 0 to infinity, rather than -infinity to infinity since p(t) doesn't exist before t = 0. Is that right ?

    (Original post by Kernel Rev)
    ...
    (b) Assuming that the cars are supplied at a constant rate, what is the probability density ?sup(t) that a given car will be supplied t years after the start of the program ?

    Now, we let X be the random variable "times after start of program that a car is supplied". We also know that the cars are supplied at a constant rate, one every 1/1000th of a year. At time t, 1000*t cars have been delivered and for every time up and til time t it was equiprobable for each of the cars deivered up to that point to have been delivered at that time so the probability of a car being delivered at time t is 1/1000*t also the p(T less than or equal t)=1 where T is the random variable "times after start of program that a car is supplied". The p.d.f which is what you want is not the probability though. This is about as far as I can get but if you have any more suggestions im gonna keep trying for a bit!
    Ok that's the probability and, from my notes, the probability is the integral of the pdf.

    P(x1,x2) = int[x1 to x2] p(x) dx

    so does that mean that,

    p(x) = dP/dx ??

    if P = t/1000
    dP/dt = 1/1000

    Now thats a constant function, and somehow or other I thought it woul be an exponential function, for when I was to do the convolution. ???
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    Yes your right about part a) just tried it again and got <t> = 3 years not 2 as I stated before. Don't do integration by parts when your tired! I would write out my solution but I think you had it nailed and im not quite sure how to use tex.

    Yes you do the integration from 0 to infinity as t does not exist for negative values. Although expectation is defined as an integral over limits -infinty to infinity, that is just for the possibility that the random variable may take values over the whole real line, usually we just deal with a subset, in our case 0 to infinity.

    As mentioned yesterday The pobability of a random variable T taking a value between t1 and t2, P[t1<equal t <equal t2] is the integral of the p.d.f function, p(t) between the limits t1 and t2. I Incorrectly gave that the probability was t/1000 when it should be t/5 as the probability is zero when to is zero and 1 when t=5 so for 0<t<5 we get that p(t) = d[P(0<t<5)]/dt = 1/5 and when t is greater than 5 the probability is zero of course so:

    psup(t) = 1/5 , 0<equal t <equal5
    = 0 , t>5

    Also interestingly if you try the interval for the probability of a car being supplied at an exact time, not within an interval we get zero, this makes sense intuitively as the cars are continuously being delivered over a time interval. A way to check a p.d.f is correct is to do the integral form -infinity to infinity of the pd.f, or in our case 0 to infinity for reasons explained earlier and you should get 1 as the probability over all possible values the R.V can take on is 1 (certain) of coursed. Doing this we get int [0,infinity] = int [0,5] + 0 = 5*1/5=1 so it works!

    Looking at part c you mention using a convolution method, Im not quite sure what you mean, I have never come across convolutions before but it could just be a different term for something I have come across. I shall try and have a look at part c) and see what I can come up with. It does seem that the pdf should be a little more exciting than a constant function but I have a feeling that an exponential may come from having to deal with rates of change and so using an ODE:

    dN/dt = constant*N*t/5 where the exponential will appear in the general solution, I don't know but we'll see!
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    Ok, thanks for that. That other web-site gave the same pdf for psup().
    I thought it would be anything but a constant function, (just a hunch/hope, that's all) but I'll go do my convolution now, using that, and see what I get.

    Edit: there's an comment on convolutions on that other site. post #13.
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    Finally gottit!! :biggrin: :biggrin:
 
 
 
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