Question

The cdf for X (=measurement error) of Exercise 3 is:

Answer 1

F(x) =

0 x < -2
\[\frac{ 1 }{ 2 } + \frac{ 3 }{ 32 }(4x - \frac{ x^3 }{ 3 }) -2 \le x < 2\]
1 2≤x

a) Compute P(X < 0)
b) Compute P(-1 < X < 1)
c) P(0.5 < X)
d) Find the density function

Answer 2

So I think I might've found the correct density function:
\[\frac{ 3 }{ 8 }(1 - \frac{ x^2 }{ 4 }) -2 \le x < 2\] and 0 otherwise

But beyond that I don't know how to start. would P(X<0) end up as 1 - F(0) = 0.5? And would P(-1 < X < 1) = F(1) - F(-1) = 0.6875?