1. ## Probability question

Can someone give me a hint on how to calculate P(|Z|∈[3/2 ; 5/2])?

Y(t)=tZ^2; t=4
Z~U(-2;2)
U - continuous uniform distribution.

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2. ## Re: Probability question

So the problem is to find $P(|Z|\in [3/2, 5/2])$? That has nothing to do with y.

Z is the uniform distribution from -2 to 2 so have value Z(z)= 1/4 for all z for all z between -2 and 2. $|Z|\in [3/2, 5/3]$ is the same as $Z\in [-2, -3/2]\cup [3/2, 2]$. Those two intervals are disjoint so that is $P(Z\in [-2, -3/2]+ P(Z\in [3/2, 2])$. And each of those is just 1/4 times the length of the interval.

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