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Show that for every $p$, $0\leq p\leq 1$, the function $f(x)$ = $p*sin(x) +(1-p)*cos(x)$, $0\leq x \leq \pi/2 $, and $f(x)=0$ otherwise, is a density function. Find its CDF and use it to find all the medians.

I was able to prove it a density function and also was able to get the CDF which is $(p +\sin(x) -p(\sin(x) + \cos(x))$ ; $0\leq x \leq \pi/2$ and $1$ for $ x\geq \pi/2$ and $zero$ elsewhere. (I hope I am correct)

I don't understand the last part. What do they mean by finding all the medians?

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    $\begingroup$ It means to derive a formula that is valid for any $p$, like a median$(p)$ function. You know that the median of Uniform$(a,b)$ is $(b-a)/2$. Do the same for your situation. $\endgroup$ – Dave Jan 11 at 23:56
  • $\begingroup$ You probably should add the self study tag. $\endgroup$ – Michael R. Chernick Jan 12 at 0:04
  • $\begingroup$ I still can't figure it out. If I put it equal to 0.5, I have an equation with two variable p and x. How can I solve this? $\endgroup$ – Leaderboard281923 Jan 12 at 0:10
  • $\begingroup$ Solve for $x$ to get the value that is the median. Write $x$ as a function of $p$. And, yes, self-study rag, please. $\endgroup$ – Dave Jan 12 at 0:24
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    $\begingroup$ Cross-posted at math.stackexchange.com/questions/3505774/…. $\endgroup$ – JimB Jan 12 at 3:27
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First median is defined where $F(x)=0.5$. Right now you have a CDF that is defined in terms of $p$ and $x$ and it is possible to define the median in terms of $p$(I did not check if your CDF is correct or not since it is your task). You can think of $p$ as a parameter and thus no need to worry if you have a definition that have $p$ in it

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  • $\begingroup$ I can't solve the equation when I am equating it to 0.5 I tried substitution $sin$$ x=t$ but still nothing. $\endgroup$ – Leaderboard281923 Jan 12 at 19:36

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