Suppose that f x 15x2 for 1 x 1 and f x 0 otherwise Dete

Suppose that f (x) = 1.5x² for -1 < x < 1 and f (x) = 0 otherwise. Determine the following probabilities. Round your answers to three decimal places (e.g. 98.765).

(a)P(0 < X)
(b)P(0.5 < X)
(c)P(-0.5 X 0.5)
(d)P(X<-2)
(e)P(X < 0 or X>-0.5)
(f) Determine x such that P(x < X) = 0.05.

Solution

f(x) = 1.5x²

The probabilities would be the integral of the given function from -1 to 1

a) P(X > 0)

= Integral from 0 to 1 of f(x)

= (1/2)x³

= 1/2 - 0

= 0.5

b)

P(X >? 0.5)

would be done in the sme way except the limits changing to { 0.5 to 1]

=- (1)³/2 - (0.5)³ /2

= 0.4375

c)

P(-0.5 < X < 0.5)

The limits become [-0.5 , 0.5]

= (0.5)³ /2 - (-0.5)³/2

= 0.125

d)

P(X< -2) = 0 since value of f(x) = 0 in that region

e)

P(X < 0 or X>-0.5 ) = P(X <0) + P(X> -0.5) - P ( -0.5 < X < 0)

= 0.5 +0.5625 - 0.0625

= 1

f) Let the value be \'a\'

Then, inmtegral from -1 to a will = 0.05

a³ - (-1)³ = 0.05 * 2

a³ + 1 = 0.10

a = (-0.90)^1/3

= -0.965

Hope this helps. Ask if you have doubts.