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f(x)= 2(x-1)/3 in the interval 1<x<2 and 4-x/3 in the intervals 2<x<4 and 0 otherwise there are 3 f(x) and I am familiar with finding the unknown variable k when there is one f(x) but in this problem there is 2 main ones the 0 is useless but I don't know which equation to fill in the numbers I tries both but they were wrong the answers spouse to be 2 but I don't know how to get it . Any help is really appreciated! This is in the topic continuous random variables in statistics and probbaility.enter image description here enter image description here

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  • $\begingroup$ Could you please ask the question in a different way? Where did the function come from? Where does the randomness come in? $\endgroup$ – rostader Jan 31 at 10:58
  • $\begingroup$ Theose are the functions there are two functions 2(x-1)/3 in the interval 1<x<2 and 4-x/3 in the intervals 2<x<4 that's what is given to me at the top of the question it is a 3 part question I dont jhave a teacher to ask so I am very confused $\endgroup$ – mary james Jan 31 at 11:02
  • $\begingroup$ OKay... but where does the probability come in? Is f a density function? $\endgroup$ – rostader Jan 31 at 11:05
  • $\begingroup$ I added pictures on how the question looks maybe that will be more clear yes it is a density function $\endgroup$ – mary james Jan 31 at 11:07
  • $\begingroup$ I whent throught it again and its PDF but what does that change $\endgroup$ – mary james Jan 31 at 11:21
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Notice that $f(1)=0$ and $f(2)=\frac{2}{3}$. Hence for $1<x\leq2$, the region under the graph of $y=f(x)$ is a triangle with area $\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$. Hence $k=2$.

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  • $\begingroup$ How did you get the f(2) $\endgroup$ – mary james Jan 31 at 11:13
  • $\begingroup$ replace $x$ with $2$ in both expressions. They both give you $\frac{2}{3}$, and assuming that the $f$ is continuous, that means that $f(2)=2$. $\endgroup$ – t-tough Jan 31 at 11:14
  • $\begingroup$ But how do you know to replace it with 2 not 1 or 3 or 4? $\endgroup$ – mary james Jan 31 at 11:15
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    $\begingroup$ If $k>2$ then we would find that out because the area under the first bit of $f$ would be less than $\frac{1}{3}$, so in order to make up the area to $\frac{1}{3}$, we would have to include some of the other bit of $f$. $\endgroup$ – t-tough Jan 31 at 11:28
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    $\begingroup$ you're very welcome :) $\endgroup$ – t-tough Jan 31 at 11:43
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For $x\in[1,\,2]$, the CDF is $\int_1^xf(t)dt=\tfrac13(x-1)^2$, so $P(X\le2)=\tfrac13$. You needn't consider the CDF on $(2,\,4]$, as clearly $k=2$.

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  • $\begingroup$ I am sorry but i am confused how did you get these numbers? $\endgroup$ – mary james Jan 31 at 11:14
  • $\begingroup$ @maryjames The support is $[1,\,4]$, so I just evaluated the CDF at $x=2$, where the PDF's formula changes, so I could see whether $k<2$, $k=2$ or $k>2$. $\endgroup$ – J.G. Jan 31 at 11:16
  • $\begingroup$ But when I replace x on both the equations it equals with 2 it equals to 2/3 $\endgroup$ – mary james Jan 31 at 11:17
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    $\begingroup$ @maryjames Well, $P(X\le k)=\int_1^kf(t)dt$, so the question at hand is about the CDF. $\endgroup$ – J.G. Jan 31 at 11:34
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    $\begingroup$ @marynames In general $P(X\le k)$ is the CDF at $k$. $\endgroup$ – J.G. Jan 31 at 11:47
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So if you check this $f$ given would be a density function for a random variable $X$ What this would mean $$\int_1^4 f(x)dx = 1$$ So what you want is to find a $k$ s.t. $$\int_1^k f(x)dx \le 1/3$$ Now the form of $f$ is given. So if $f$ was defined in the same way over the entire domain then you could replace $f$ with that definition and performed an integration with $k$ being solved as an unknown. However as $f$ is defined over 2 intervals you should take care and take 2 cases. One when $k$ belongs to the first and the other when $k$ belongs to the second and proceed in the manner similar to if there was only a single equation. Ofcourse since $f$ is a density function the value of $k$ would be unique. So you have to eliminate one of the choices of $k$ you get from considering the 2 equations using the definition of the function.

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  • $\begingroup$ aw okay thank you:) $\endgroup$ – mary james Jan 31 at 11:24

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