6 Suppose fx 15x2 for 1 SolutionIntegrating fx Integral fx

6. Suppose f(x) = 1.5x^2 for - 1

Solution

Integrating f(x),

Integral [f(x) dx] = 0.5x^3

Thus,

a)

P(-0.5<x<0.5) = 0.5(0.5)^3 - 0.5(-0.5)^3 = 0.125 [ANSWER]

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b)

As for x < -1, f(X) = 0, and -2 < -1, then

P(x < -2) = 0 [answer]

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c)

Obviously, X has to be between -1 and 1.

Thus,

P(x<X) = 0.5X^3 - 0.5(-1)^3 = 0.05

0.5X^3 = -0.45

X = -0.96549 [ANSWER]