Question

The probability distribution function of a random variable X with pos- sible values in the range 0 〈 x 〈 oo is f(x) exp(-ax). Determine (a) the numerical value of α (b) P(X 〉 2) (c) Pi 〈 x 〈 3) (d) the value x such that P(X 〈 x) 0.1.

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Answer #1

a) \int_{0}^{\infty }f(x)dx = 1

\Rightarrow \int_{0}^{\infty }e^{-\alpha x}dx = 1

\Rightarrow [\frac{-e^{-\alpha x}}{\alpha }]_{0}^{\infty } = 1

\Rightarrow \frac{1}{\alpha }= 1

\Rightarrow \alpha = 1

The pdf of X is

f(x) = e^{-x},x>0

The cumulative distribution function of X is

F(x) = \int_{0}^{x}f(t)dt =\int_{0}^{x} e^{-t}dt = [-e^{-t}]_{0}^{x}= 1 - e^{-x}

\Rightarrow F(x) = 1 - e^{-x}, x>0

b) P(X>2)= 1 - P(X\leq 2) = 1 - F(2)= 1 - (1 - e^{-2})= e^{-2} = 0.135335

c) P(1<X<3) = P(X<3) - P(X<1) = F(3) - F(1)

\Rightarrow P(1<X<3) = F(3) - F(1)

\Rightarrow P(1<X<3) = (1-e^{-3}) - (1-e^{-1}) = e^{-1}-e^{-3} = 0.318092

\Rightarrow P(1<X<3) = 0.318092

d) To find x such that

P(X<x) = 0.1

\Rightarrow F(x) = 1-e^{-x} = 0.1

\Rightarrow e^{-x} =1 - 0.1 = 0.9

\Rightarrow -x = ln( 0.9) = -0.105361

\Rightarrow x = 0.105361

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