# Find the value of k such that P(X<k)=1/3 With additional information in description

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.  • Could you please ask the question in a different way? Where did the function come from? Where does the randomness come in? – rostader Jan 31 at 10:58
• 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 – mary james Jan 31 at 11:02
• OKay... but where does the probability come in? Is f a density function? – rostader Jan 31 at 11:05
• I added pictures on how the question looks maybe that will be more clear yes it is a density function – mary james Jan 31 at 11:07
• I whent throught it again and its PDF but what does that change – mary james Jan 31 at 11:21

Notice that $$f(1)=0$$ and $$f(2)=\frac{2}{3}$$. Hence for $$1, the region under the graph of $$y=f(x)$$ is a triangle with area $$\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$$. Hence $$k=2$$.

• How did you get the f(2) – mary james Jan 31 at 11:13
• replace $x$ with $2$ in both expressions. They both give you $\frac{2}{3}$, and assuming that the $f$ is continuous, that means that $f(2)=2$. – t-tough Jan 31 at 11:14
• But how do you know to replace it with 2 not 1 or 3 or 4? – mary james Jan 31 at 11:15
• 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 $f$. – t-tough Jan 31 at 11:28
• you're very welcome :) – t-tough Jan 31 at 11:43

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 $$k=2$$.

• I am sorry but i am confused how did you get these numbers? – mary james Jan 31 at 11:14
• @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 $k>2$. – J.G. Jan 31 at 11:16
• But when I replace x on both the equations it equals with 2 it equals to 2/3 – mary james Jan 31 at 11:17
• @maryjames Well, $P(X\le k)=\int_1^kf(t)dt$, so the question at hand is about the CDF. – J.G. Jan 31 at 11:34
• @marynames In general $P(X\le k)$ is the CDF at $k$. – J.G. Jan 31 at 11:47

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.

• aw okay thank you:) – mary james Jan 31 at 11:24