Suppose that f (x) = 1.5x2 for -1 < x < 1 and f (x) = 0 otherwise. Determine the following probabilities. 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.

a) P(0 < X) = 0.5

b) P(0.5 < X) = 0.4375

c) P(-0.5 = X = 0.5) = 0.125

d) P(X< -2) = 0

e) P(X < 0 or X>-0.5) = 1

f) value of x is -0.9654

Step-by-step explanation:

Given that;

f(x) = 1.5x²

Now. the probabilities would be the integral of the given function from (-1 to 1)

a) P(0 < X)

P(X > 0)

⇒ Integral from 0 to 1 of f(x)

= (1/2)x³

= 1/2 - 0

= 0.5

Therefore; P(0 < X) = 0.5

b) P(0.5 < X)

P(X > 0.5)

this will also be done in the same way except that, the limits changes to { 0.5 to 1]

= (1)³/2 - (0.5)³ /2  

= 0.4375

Therefore; P(0.5 < X) = 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

Therefore; P(-0.5 = X = 0.5) = 0.125

d) P(X<-2)

since value of f(x) = 0 in that region

Therefore P(X< -2) = 0

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

Therefore; P(X < 0 or X>-0.5) = 1

f) Determine x such that P(x < X) = 0.05.

Let the value be 'a'

Then, integral from -1 to a will = 0.05

a³ - (-1)³ = 0.05 × 2

a³ + 1 = 0.10

a³ = 0.10 - 1

a³ = -0.9

a = ∛-0.9  

a = -0.9654

Therefore, value of x is -0.9654

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.