Showing a random Variable has a continuous uniform distribution

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To show that Z has a continuous uniform distribution over the interval (a,b), start with the transformation Z=(b-a)(x-θ)+(1/2)(a+b). By letting y = x-θ, it is established that y is uniformly distributed over the interval (-1/2, 1/2). Since Z is a linear transformation of y, it follows that Z is also uniformly distributed. The endpoints of Z can be determined by substituting the endpoints of y, resulting in Z=a when y=-1/2 and Z=b when y=1/2. This confirms that Z maintains a continuous uniform distribution over the specified interval.
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f(x)=1, θ-1/2 ≤ x ≤ θ+1/2

Given that Z=(b-a)(x-θ)+(1/2)(a+b) how would you show that Z has a continuous uniform distribution over the interval (a,b)?
Any help would be much appreciated.
 
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Let y = x-θ. Then y is uniform over the interval (-1/2,1/2). Since Z is linear in y, it is also uniformly distributed.
Put in the endpoints of y to get the endpoints for the Z interval. At y = -1/2, Z = a, while at y = 1/2, Z = b.

If you want to do a little work, start with P(Z < z) and transform it into P(Y < y) or P(X < x).
 
Ohh yeah that makes a ton of sense. Idk what I was thinking. Thanks!
 
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