Distribution Function Calculations for Given Probabilities

Leper Messiah
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1.Given f(-1) = 0.15, f(1) = 0.29, f(3) = 0.21, and f(5) = 0.35, find P(X ≤ 1)

2.Given f(-1) = 0.15, f(1) = 0.29, f(3) = 0.21, and f(5) = 0.35, find F(1)

3.Given f(-1) = 0.15, f(1) = 0.29, f(3) = 0.21, and f(5) = 0.35, find P(-1 < X ≤ 3)

and the rest using this distribution function
F(x) = [0 for x < 1; 1/3 for 1 <= x < 4; 1/2 for 4 <= x < 6; 5/6 for 6 <= x < 10; 1 for x >= 10]

4.find P(2 ≤ X < 6)

5.find P(X = 5)

6.find P(X ≤ 6)

I know this is a lot, but help is really needed and I would be much appreciative!
 
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You also need to read the rules of this forum (which you should have done already): show what you have done already on this problem so we will know what hints you need.

It also would be a good idea to clearly state the problem. Are you saying that the first 3 problems are discrete probabilities with 5 outcomes? And that the last problems are continuous probabilities? And what is the difference between f(X) and F(X)?
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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