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The C.D.F. (cumulative density function) of the random variable X is given by,
F(x)= [0, x<0
x/2, 0<=x<1
2/3, 1<=x<2
11/12, 2<=x<3
1, 3<=x]
a) Find P(X>1/2)
b) Find P(2<X<=4)
c) Find P(X=1)
Does anyone know how to approach this?
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Recall the definition (note that X is a random variable and x is the argument to the function):
the probability P[X <= x] = F(x)
We know F(x).
a) P[X > 1/2] = 1 - P[X <= 1/2]
now use the definition of F(x):
P[X <= 1/2] = F(1/2) = (1/2)/2 = 1/4
since this is F(x) for 0 <= x < 1
So now,
P[X > 1/2] = 1 - P[X <= 1/2] = 1 - 1/4 = 3/4
I don't quite remember how to do the others at this moment, hopefully other posters can help!
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(b)
Here I think you need the formula:
P[a < X <= b] = F(b) - F(a)
so
P[2 < X <= 4] = F(4) - F(2) = 1 - 11/12 = 1/12
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