Given the CDF of a random variable Given the CDF of a random

Given the CDF of a random variable

Given the CDF of a random variable Given the CDF of random variable X, find P(X GT 1/2). What is the value of a such that P(X LE a) = .9 Determine the Probability Density Function (PDF) from the CDF. What type of distribution is this?

Solution

A)

P(x>1/2) = 1 - F(1/2) = 1 - (1/2 + 1)/2 = 0.25 [answer]

B)

F(a) = 0.9 = (x + 1)/2

Thus,

x = 0.8 [answer]

C)

The pdf is the derivative of F(x).

Thus,

f(x) = 0, x<-1, x>=1
1/2, -1<=x<1 [ANSWER]

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d)

This is a uniform distribution, as f(x) is constant over an interval.