Computing probabilities from a piecewise CDF

[sneaky666]

Homework Statement

suppose Fy(y)=y^3 for 0<=y<1/2 and Fy(y)=1-y^3 for 1/2<=y<=1. Compute these.

1.
P(1/3<Y<3/4)
2.
P(Y=1/3)
3.
P(Y=1/2)

Homework Equations

The Attempt at a Solution

Is this right for the 1. ?
P(1/3<Y<3/4)
P(1/3<Y<1/2) + P(1/2<=Y<3/4)
( Fy(1/2-) - Fy(1/3) ) + ( Fy(3/4-) - Fy(1/2-) )
Fy(1/2-) - Fy(1/3) + Fy(3/4-) - Fy(1/2-)


Now I have to use
Fy(y)=y^3 for 0<=y<1/2 and Fy(y)=1-y^3 for 1/2<=y<=1

1/8 - 1/27 + 37/64 - 1/8
=935/1728

for 2 i am getting 1/27
for 3 i am getting 7/8

I think there's something wrong, but I don't know what.

 

[vela]

Is F_y supposed to be the cumulative distribution function? If it is, you must have written down the wrong function because the cdf should be an increasing function. If it's supposed to be a probability density function, there's also a problem because the area under the curve from -∞ to +∞ isn't equal to 1.

 

[sneaky666]

the exact question was:

Suppose Fy(y)=y^3 for 0<=y<1/2, and Fy(y)=1-y^3 for 1/2<=y<=1. Compute each of the following:
a) P(1/3<Y<3/4)
b) P(Y=1/3)
c) P(Y=1/2)

the section name is cumulative distribution function
chapter name is random variables and distribution

 

[vela]

I think you need to go ask your instructor about this problem because it doesn't make sense.