Find k if the following function represent p.d.f. of r.v. X.

f (x) = kx (1 – x), for 0 < x < 1 and = 0 otherwise, Also find P `(1 /4 < x < 1 /2) , P (x < 1 /2)`.

#### Solution

Since, the function f is the p.d.f. of X,

` int_(-∞)^∞ f (x) dx` = 1

∴ ` int_(-∞)^0 f (x) dx`+` int_(0)^1 f (x) dx`+` int_(1)^∞f (x) dx` = 1

∴ 0+` int_(0)^1kx (1-x) dx + 0` = 1

∴ k` int_(0)^1 (x - x^2) dx = 1`

∴ k `[x^2/2-x^3/3]_0^1` = 1

∴ k `(1/2 - 1/3-0)` = 1

∴ `k/6` = 1

∴ k = 6

P `(1 /4 < x < 1 /2)` = ` int_(1/4)^(1/2) f (x) dx`

= ` int_(1/4)^(1/2) kx( 1-x) dx`

= k` int_(1/4)^(1/2) (x-x^2) dx`

= `6[x^2/2 - x^3/3]_(1/4)^(1/2)`..........[∵ k = 6]

= `6 [(1/8-1/24)-(1/32-1/192)]`

= `6[2/24-5/192]`

= `6 (11/192)`

∴ P `(1 /4 < x < 1 /2)` = `11/32`

P `(x < 1 /2)` = ` int_(0)^(1/2) f (x) dx`

= 0 +` int_(0)^(1/2) kx(1 - x) dx`

= k` int_(0)^(1/2) (x - x^2) dx`

= k`[x^2/2-(x^3)/3]_0^(1/2)`

=`k [1/8-1/24-0]`

= `k (2/24)`

= `6 (1/12) `............[∵ k = 6]

∴ P `(x < 1/2)`

= `1/2`