1.

Find k if the function f(x) is defined by `f(x)=kx(1-x),"for "0ltxlt1` =0 , otherwise, is the probability density function (p.d.f.) of a random varible (r.v) X. Also find P `(Xlt(1)/(2))`

Answer» Since the givne function is p.d.f., we get
`int_(0)^(1)f(x).dx=1`
`:.int_(0)^(1)kx(1-x).dx=1`
`:.int_(0)^(1)(x-x^(2)).dx=1`
`:.k[(x^(2))/(2)-(x^(3))/(3)]^(1)=1`
`:.(1^(2))/(2)-(1^(3))/(3)=(1)/(k)`
`:.(1)/(6)=(1)/(k)`
`:.k=6`
Now `p[xlt(1)/(2)]=int_(0)^(1//2)f(x)dx`
`=int_(0)^(1//2)kx(1-x).dx`
`=int_(0)^(1//2)6x(1-x).dx`
`int_(0)^(1//2)(6x-6x^(2)).dx`
`=[(6x^(2))/(2)-(6x^(3))/(3)]_(0)^(1//2)`
`=[3x^(2)-2x^(3)]_(0)^(1//2)`
`=3((1)/(2))^(2)-2((1)/(2))^(3)`
`=(3)/(4)-(1)/(4)=(2)/(4)=(1)/(2)`
`p[xlt(1)/(2)]=(1)/(2)`


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