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    This is the only thing I have trouble with in S2...my exam is on wednesday and I really struggle with this chapter, never got exalained properly. If you could help with through this question maybe I'll undersand it more?

    The waiting time, T minutes, before being served at a local newsagent can be modelled by a continuous random variable with probability density function:

    f(t) = 3/8 t^2 o<t<1
    1/16 (t+5) 1<t<3
    0 Otherwise

    Show that the cumulative distribution function for 1<t<3 is given by

    F(t)= 1/32 (t^2 + 10t - 7)

    Hence find the median waiting time.
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    Do you know what you're supposed to do? An attempt?
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    (Original post by SLB1904)
    Show that the cumulative distribution function for 1<t<3 is given by

    F(t)= 1/32 (t^2 + 10t - 7)

    Hence find the median waiting time.
    Here's my guess: the cumulative distribution function is probably the area under the curve, and the median is probably this area divided by the length of the time-interval over which you integrated. I've never done any probability though so I may be completely wrong..

    (Not an "explanation" sorry, but I posted anyway because I'm bored.)
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    (Original post by silent ninja)
    Do you know what you're supposed to do? An attempt?
    I integrated from 0 <t< 1 to get 1/8 t^3

    then integrated from 1 < t < 3 (using reverse chain rule)

    so I got F(1) + 1/32 (t+5)^2 (F(1)= 1/8)

    and i have no idea what to do from here.
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    How did they get the -7?
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    (Original post by SLB1904)
    I integrated from 0 <t< 1 to get 1/8 t^3

    then integrated from 1 < t < 3 (using reverse chain rule)

    so I got F(1) + 1/32 (t+5)^2 (F(1)= 1/8)

    and i have no idea what to do from here.
    so 0 <t< 1 you get 1/8 t^3. If you plug those limits in you get 1/8.

    Next 1/32 (t+5)^2 + c --you need a constant. But you already know if you plug in t=1 the area should be 1/8 so find c. Then simplify.
    Note: it is a continuous distribution so the value t=1 can be in 'both' functions (continuous overlap)
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    (Original post by silent ninja)
    so 0 <t< 1 you get 1/8 t^3. If you plug those limits in you get 1/8.

    Next 1/32 (t+5)^2 + c --you need a constant. But you already know if you plug in t=1 the area should be 1/8 so find c. Then simplify.
    Can I ask you something?

    Would it also be correct to say that the total (cumulative) probability at t=3 must equal 1? I also get -7 this way, but I don't know if that is coincidence (since they don't specify that all customers get served before t=3).

    Edit: oh wait they actually do specify it. So I guess that is correct.
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    (Original post by llys)
    Can I ask you something?

    Would it also be correct to say that the total (cumulative) probability from t=0 to t=3 must equal 1? I also get -7 this way, but I don't know if that is coincidence (since they don't specify that all customers get served before t=3).
    Well F(3) is supposed to be 1 because the total area is always 1; they dont need to say this it's just a fact of probability otherwise the initial function they gave you is not a pdf.
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    (Original post by silent ninja)
    Well F(3) is supposed to be 1 because the total area is always 1; they dont need to say this it's just a fact of probability otherwise the initial function they gave you is not a pdf.
    Right. So why do you use the integration constant c and the condition at t=1, F=1/8, instead of using no integration constant (since by integration from 1 to 3 the constant c would fall out anyway?). Why don't you just normalize the area (F(3)=1) ?
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    F(3)=1 means  \int_0^1 + \int_1^3 (for each pdf function) Yes that's fine and all the constants cancel out and it looks like everything is fine. But think of it this way: for 1<t<3 if I plug in any t=x value into 1/32(t+5)^2 does it actually take into account the area between 0<t<1 ? After all this integral is only for the range 1<t<3. If it's a CDF it should include the entire area below the value t=x.
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    (Original post by silent ninja)
    F(3)=1 means  \int_0^1 + \int_1^3 (for each pdf function) Yes that's fine and all the constants cancel out and it looks like everything is fine. But think of it this way: for 1<t<3 if I plug in any t=x value into 1/32(t+5)^2 does it actually take into account the area between 0<t<1 ? After all this integral is only for the range 1<t<3. If it's a CDF it should include the entire area below the value t=x.
    I don't think that's what I'm asking though. (I'm not sure.) What I would do to determine F(t) in the interval 1<t<3 is this:

    area normalisation:
    (1/8+1/32(3+5)²-1/32(1+5)²)/N==1,

    where then later
    F(t) for 1<t<3 is equal to
    F(t)=(1/8+1/32(t+5)²-1/32(1+5)²)/N

    Would this be right or wrong? Thanks. :-)
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    (Original post by llys)
    I don't think that's what I'm asking though. (I'm not sure.) What I would do to determine F(t) in the interval 1<t<3 is this:

    area normalisation:
    (1/8+1/32(3+5)²-1/32(1+5)²)/N==1,

    where then later
    F(t) for 1<t<3 is equal to
    F(t)=(1/8+1/32(t+5)²-1/32(1+5)²)/N

    Would this be right or wrong? Thanks. :-)
    That first bold bit totals 1 though?

    In a roundabout way you're doing the same thing.
    The second bold bit but just means \int_1^t \frac{1}{16}(t+5) dt + \frac{1}{8} which is what we're after for this range, yes. Basically the same method.
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    (Original post by silent ninja)
    That first bold bit totals 1 though?
    Yes - it just so happens that here N=1. The difference is just the way of approaching it. I don't know if the area is always automatically already normalized in these exercises. I know the total probability always has to equal 1, but I don't know if that normalization is always already inbuilt in the exercise or if the student sometimes has to do it. This is why I first thought of using this approach..

    Anyway I see now. :woo:
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    (Original post by llys)
    Yes - it just so happens that here N=1. The difference is just the way of approaching it. I don't know if the area is always automatically already normalized in these exercises. I know the total probability always has to equal 1, but I don't know if that normalization is always already inbuilt in the exercise or if the student sometimes has to do it. This is why I first thought of using this approach..

    Anyway I see now. :woo:
    No you dont need to normalise anything. You are just adding areas and making sure they are consistent for different ranges. The more direct approach is just to write something like that last integral first off and this is usually how it's written in mark schemes.
 
 
 
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