Suppose the random variable X has pdf given by f x 14 x3 0

Suppose the random variable X has pdf given by f( x) = 1/4 x^3 , 0 x 2

a. Find P( X 1)

b. Find P(X 1)

c. Find the value of c such that P( X c) = 0.05

Solution

Here,

F(x) = Integral [f(x) dx] = Integral [1/4 x^3 dx]

F(x) = x^4 / 16, 0<=x<=2

Thus,

a)

P(x<=1) = F(1) = 1/16 = 0.0625 [answer]

b)

P(x>=1) = 1 - F(1) = 1-1/16 = 0.9375 [answer]

c)

If

P(x<=c) = 0.05, then

F(c) = c^4/15 = 0.05

c^4 = 0.75

c = 0.930604859 [answer]