Advertisement

Advertisement

Sum

The p.d.f. of a continuous r.v. X is given by

f (x) = `1/ (2a)` , for 0 < x < 2a and = 0, otherwise. Show that `P [X < a/ 2] = P [X >( 3a)/ 2]` .

Advertisement

#### Solution

P (`X< a /2`) = ` int_(-∞)^(a/2) f (x) dx`

=` int_(-∞)^0 f (x) dx+ int_(-∞)^(a/2) f (x) dx`

= 0+ ` int_(0)^(a/2) 1/(2a) dx`

= `1/(2a) int_(0)^(a/2) 1 dx = 1/(2a)[x]_0^(a/2)`

= `1/(2a) [a/2-0] = 1/4` .......(1)

P `[X >( 3a)/ 2] = int_((3a)/2)^(∞) f(x) dx`

= `int_((3a)/2)^(*2a) f(x) dx + int_(2a)^(∞) f(x) dx`

= `int_((3a)/2)^(2a) 1/(2a) dx + 0`

= `1/(2a) int_((3a)/2)^(2a) 1 dx = 1/(2a)[x]_((3a)/2)^(2a)`

= `1/(2a)[2a-(3a)/2] = 1/(2a)(a/2) = 1/4` ......(2)

From (1) and (2), we get

`P [X < a/ 2] = P [X >( 3a)/ 2]`.

Is there an error in this question or solution?