Find cumulative distribution

m.ooo

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Dec 1, 2021
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Okay so I have this exam prep but I dont really understand how I am supposed to solve this. Can anyone help?as.png
 
Seriously, I don't know why I stuck on that question for so long. Thanks! :)
*The answer is 4/5
No the answer is 410\dfrac{4}{10} they must add to 11
 
F(a)={0if a<01/10if 0a<55/10if 5a<108/10if 10a<151if a15F(a) = \begin{cases} 0 &\text{if } a<0 \\ 1/10 &\text{if } 0\le a <5 \\ 5/10 &\text{if } 5\le a <10 \\ 8/10 &\text{if } 10\le a <15 \\ 1 &\text{if } a\ge15 \\ \end{cases}I don't see how the answer can be 4/5.
Well, my bad, I had in mind that F(a)=8/10 if 15<=a<20 .
 
F(a)={0if a<01/10if 0a<55/10if 5a<108/10if 10a<151if a15F(a) = \begin{cases} 0 &\text{if } a<0 \\ 1/10 &\text{if } 0\le a <5 \\ 5/10 &\text{if } 5\le a <10 \\ 8/10 &\text{if } 10\le a <15 \\ 1 &\text{if } a\ge15 \\ \end{cases}I don't see how the answer can be 4/5.
Correction:
F(a)={0if a<51/10if 5a<105/10if 10a<158/10if 15a<201if a20F(a) = \begin{cases} 0 &\text{if } a<5 \\ 1/10 &\text{if } 5\le a <10 \\ 5/10 &\text{if } 10\le a <15 \\ 8/10 &\text{if } 15\le a <20 \\ 1 &\text{if } a\ge20 \\ \end{cases}Therefore, F(17)=8/10=4/5F(17)=8/10=4/5
 
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The F(17) is 8/10 -> 4/5. I assume you're talking about the p(10).
My bad. Yes that was p(10)p(10). So F(17)=810F(17)=\dfrac{8}{10}
 
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