How can i solve this probability problem?

Charles Borges said:
View attachment 21160
Click to expand...
What they are asking for is closely related to what the CDF does: The CDF is defined by [MATH]F(x) = P(X\le x)[/MATH]. You want to find the value of y such that [MATH]P(X>y) = 0.19[/MATH]. Do you see the relationship?

(I would have said "find" rather than "define" there. As it is, it doesn't ask you a question!)
 
Subhotosh Khan said:
What is the relationship between

Cumulative distribution function and​
​
Probability​

Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:

https://www.freemathhelp.com/forum/threads/read-before-posting.109846/#post-486520

Please share your work/thoughts about this problem.
Click to expand...

Well, I tried it like this:

P (X > y) = 0.19 => F (y) = 0.19 => 0.25 * y² = 0.19, so I made the substitutions and then took the square root of y² and found y = 0.774596669241483 . But looks like it's incorrect, so I asked for help.

pka said:
Given the posted attached image, it is hard to know the question.
I assume that we are to find the \(y\in (0,2)\) such that \(F(y)>0.19\)
Can you do that?

If that is not what you understand the question to be, please correct me.
Click to expand...
I need find y that make P (X>y) exactly 0.19.

Dr.Peterson said:
What they are asking for is closely related to what the CDF does: The CDF is defined by [MATH]F(x) = P(X\le x)[/MATH]. You want to find the value of y such that [MATH]P(X>y) = 0.19[/MATH]. Do you see the relationship?

(I would have said "find" rather than "define" there. As it is, it doesn't ask you a question!)
Click to expand...


Yes, that is exactly what was asked for. Excuse me, I translated poorly. So, I understood the problem, but I just can't seem to solve it. It seems like a simple problem, but when I put it on paper it just doesn't work
 
"y = 0.774596669241483 "?

Why did you use 0.15, rather than 0.19? Maybe you should just try that square root, again.
 
Charles Borges said:
Well, I tried it like this:

P (X > y) = 0.19 => F (y) = 0.19 => 0.25 * y² = 0.19, so I made the substitutions and then took the square root of y² and found y = 0.774596669241483 . But looks like it's incorrect, so I asked for help.

I need find y that make P (X>y) exactly 0.19.

Yes, that is exactly what was asked for. Excuse me, I translated poorly. So, I understood the problem, but I just can't seem to solve it. It seems like a simple problem, but when I put it on paper it just doesn't work
Click to expand...
But F(y) is not P(X > y); it's P(X ≤ y), as I stated. What do you really want F(y) to equal??
 
Dr.Peterson said:
But F(y) is not P(X > y); it's P(X ≤ y), as I stated. What do you really want F(y) to equal??
Click to expand...

Yes, you are correct. I realized that a moment ago. That was the problem, correcting this I managed to get the right result
So i did:

1 - F(y) = 0.19
=> 1 - (0.25 * y²) = 0.19

solving this equation i got the correct result of y

Thank you all for your help.
 
You must log in or register to reply here.
Top