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S2 CRV question help!! Watch

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    Hi!!!
    So, one of the questions on my homework looks like this:
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    and I got a = -26.4 and b = 39/55

    Just wanted to know if this is right because they're weird numbers lol thanks
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    (Original post by ltbitters)
    Hi!!!
    So, one of the questions on my homework looks like this:
    Name:  ps.png
Views: 95
Size:  16.8 KB
    and I got a = -26.4 and b = 39/55

    Just wanted to know if this is right because they're weird numbers lol thanks
    One way to check is to integrate f(x) between 1 and 0 and seeing if it is equal to 1. As F(1) = 1.
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    (Original post by SeanFM)
    One way to check is to integrate f(x) between 1 and 0 and seeing if it is equal to 1. As F(1) - F(0) should equal 1.
    oh nooo it doesnt equal zero and idk what i did wrong D:
    i tried creating simultaneous equations with E(X) and integrating to equal 1 rip
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    (Original post by ltbitters)
    oh nooo it doesnt equal zero and idk what i did wrong D:
    i tried creating simultaneous equations with E(X) and integrating to equal 1 rip
    It should equal 1 but I think that's a typo.

    Also, f(x) can't be negative so your -26.4 is a bit of a red flag but no worries

    What were your simultaneous equations?
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    (Original post by SeanFM)
    It should equal 1 but I think that's a typo.

    Also, f(x) can't be negative so your -26.4 is a bit of a red flag but no worries

    What were your simultaneous equations?
    ahhh yep it didnt equal 1 either tho so lol :lol: I had no idea f(x) couldn't be negative though so thanks for the tip there

    I had 0.6 = a(1/4 b - 1/5) and 1 = a(1/3 b - 1/4) and then multiplied them both so they equalled 6
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    (Original post by ltbitters)
    ahhh yep it didnt equal 1 either tho so lol :lol: I had no idea f(x) couldn't be negative though so thanks for the tip there

    I had 0.6 = a(1/4 b - 1/5) and 1 = a(1/3 b - 1/4) and then multiplied them both so they equalled 6
    Yeah, f(x) is the probability of whatever it is being x, so if it's negative.. then you have a negative probability :hide:

    How did you get those questions though? I'm sure you solved them correctly but maybe they aren't the right ones to use..

    Given that the mean is 600 (how can you find the mean in terms of a and b, using properties of a CRV?) and F(upper limit, in this case 1) where F is the CDF is equal to 1.. you should be able to get 2 equations.
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    hoo boy i think i found what i did wrong lmaooo i divided 10 by 6 and got 0.6 :lol:
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    (Original post by SeanFM)
    Yeah, f(x) is the probability of whatever it is being x, so if it's negative.. then you have a negative probability :hide:

    How did you get those questions though? I'm sure you solved them correctly but maybe they aren't the right ones to use..

    Given that the mean is 600 (how can you find the mean in terms of a and b, using properties of a CRV?) and F(upper limit, in this case 1) where F is the CDF is equal to 1.. you should be able to get 2 equations.
    ohh i didnt use a cdf :0
    i just set E(X) to 0.6 (x is in the thousands or litres and the mean is 600 litres) and then integrated x.(abx^2-ax^3) between 1 and 0 for the first equation and integrated abx^2-ax^3 between 1 and 0 and set it equal to 1 for the second??

    im not sure if my equations are right tbh but i found the most stupid mistake in my working (10/6 = 0.6 apparently) and now my answers are a=12 and b=1 which seem like a lot nicer numbers lmao
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    (Original post by ltbitters)
    ohh i didnt use a cdf :0
    i just set E(X) to 0.6 (x is in the thousands or litres and the mean is 600 litres) and then integrated x.(abx^2-ax^3) between 1 and 0 for the first equation and integrated abx^2-ax^3 between 1 and 0 and set it equal to 1 for the second??

    im not sure if my equations are right tbh but i found the most stupid mistake in my working (10/6 = 0.6 apparently) and now my answers are a=12 and b=1 which seem like a lot nicer numbers lmao
    Ah good, your method to get the equations were right then

    You can use your integrated step as well to find F(1) (substituting in a and b) and then you'll be able to check
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    (Original post by SeanFM)
    Ah good, your method to get the equations were right then

    You can use your integrated step as well to find F(1) (substituting in a and b) and then you'll be able to check
    awesome !! i think i might have this now tytyty so much for the help lol sorry for wasting time w/ a stupid mistake tho :lol:
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    (Original post by ltbitters)
    awesome !! i think i might have this now tytyty so much for the help lol sorry for wasting time w/ a stupid mistake tho :lol:
    No problem at all if you can, with questions spot if you can self-validate the answer in any way using what you know.. (for simultaneous equations in general, see if the solutions satisfy one or both equations for example) or use CDFs/PDFs etc..
 
 
 
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