Suppose Y is a continuous variable whose density gives the following probabilities:
Pr{ Y < -3 } = 0.05
Pr{ -3 < Y < -1 } = 0.15
Pr{ -1 < Y < 0 } = 0.3
Pr{ 0 < Y < 1 } = 0.3
Pr{ 1 < Y < 3 } = 0.15
Pr{ Y > 3 } = 0.05
Compute Pr{ Y > 0 } = ????
Thank you!
Clearly that would be 0.3 + 0.15 + 0.05 = 0.5
Makes sense, since both the intervals and the probabilities are symmetric.
Extra credit: what is the probability that Y=0?
That's equal to zero :) :) :)
For this question I got .45 and it was automatically marked as wrong so I was frustrated!
Thank you!
To compute Pr{Y > 0}, we need to calculate the probability of Y being greater than 0, given the provided information about its density.
First, let's visualize the given probability distribution by drawing a number line and representing the probabilities as shaded regions.
---------------------|------------------------|-----------------
-3 0 3
Based on the given information, we can conclude:
Pr{Y < -3} = 0.05 (Area to the left of -3)
Pr{-3 < Y < -1} = 0.15 (Area between -3 and -1)
Pr{-1 < Y < 0} = 0.3 (Area between -1 and 0)
Pr{0 < Y < 1} = 0.3 (Area between 0 and 1)
Pr{1 < Y < 3} = 0.15 (Area between 1 and 3)
Pr{Y > 3} = 0.05 (Area to the right of 3)
Using this information, we can find the probability of Y being greater than 0 by summing the probabilities of the regions to the right of 0:
Pr{Y > 0} = Pr{0 < Y < 1} + Pr{1 < Y < 3} + Pr{Y > 3} = 0.3 + 0.15 + 0.05 = 0.5
Therefore, Pr{Y > 0} = 0.5, which means there is a 50% probability that Y is greater than 0.