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123ertyy
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#1
Report Thread starter 9 years ago
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Hi

I found this markscheme (In S2) that i think theres seomthing wrong, could you check it please?



Here the integral of 1/3 isnt included. only 1/3.

is this the way you do it?
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lilangel890
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What's the question? It's confusing without.

Edit: I think I see what's going on. To get the final cdf for 1<x<2 you have to include F(1). When you sub 1 into 1/3x you get...
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nuodai
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Way to give us half the question :p:

I'm guessing it says something like f(x) = 1/3 in the range 0 < x < 1, since the limit on the integral changed

So, \displaystyle \int^1_0 \dfrac{1}{3}\ \mbox{d}x = \left[ \dfrac{1}{3}x \right]^1_0 = \dfrac{1}{3}

That's all they've done.
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123ertyy
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Report Thread starter 9 years ago
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(Original post by nuodai)
Way to give us half the question :p:

I'm guessing it says something like f(x) = 1/3 in the range 0 < x < 1, since the limit on the integral changed

So, \displaystyle \int^1_0 \dfrac{1}{3}\ \mbox{d}x = \left[ \dfrac{1}{3}x \right]^1_0 = \dfrac{1}{3}

That's all they've done.


Ok thats ehm, kinda embarassing :P ok thanks for your help again


(Original post by lilangel890)
What's the question? It's confusing without.

Edit: I think I see what's going on. To get the final cdf for 1<x<2 you have to include F(1). When you sub 1 into 1/3x you get...


yeap thats right

Thanks for your helpful responses
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thepieofpie
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Report 9 years ago
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It's what nuodai said, this is how you get the CDF from PDF.
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