Answer to Question #157011 in Statistics and Probability for sam21ucd

Question #157011

The probability density function of a random variable X is given by :

f(x)={kx(2-x), 0<x<2

{0, e.w

a) Find the value of k

b)Find the distribution function F(X)

c)Find P(X>3)


1
Expert's answer
2021-01-22T03:26:42-0500

(a) Probability density function of a random veriable XX is given by

f(x)=kx(2x)f(x)=kx(2-x) , 0<x<20<x<2

=0,=0, e.w

We know that for p.d.fp.d.f ,

f(x)dx=1\int _{-\infty}^{\infty} f(x)dx=1

    02kx(2x)dx=1\implies \int _ {0}^{2} kx(2-x)dx=1

    k.02(2xx2)dx=1\implies k.\int _ {0}^{2}(2x-x^2)dx=1

    k.[2.x22x33]02=1\implies k.[2.\frac{x^2}{2}-\frac {x^3}{3}]_{0}^{2}=1

    k.[483]=1\implies k.[4-\frac{8}{3}]=1

    k=34\implies k=\frac{3}{4}


(b) The distribution function of the continuous random variable XX having probability density function f(x)f(x) is given by ,

FX(x)=P(Xx)=xf(u)duF_X(x)=P(X\leq x)=\int _{-\infty}^{x}f(u) du

Therefore for the given p.d.fp.d.f , the distribution function FX(x)=xf(u)duF_X(x)=\intop_{-\infty}^{x}f(u)du

=0x34u(2u)du=\intop_{0}^{x}\frac{3}{4}u(2-u)du

=340x(2uu2)du=\frac{3}{4}\intop_{0}^{x}(2u-u^2)du

=34[u2u33]0x=\frac{3}{4}[u^2-\frac{u^3}{3}]_{0}^{x}

=14[3x2x3]=\frac{1}{4}[3x^2-x^3]


(c) P(X>3)=3f(x)dx=0P(X>3)=\intop_{3}^{\infty}f(x)dx=0 [ as f(x)=0f(x)=0 when 3<x<3<x<\infty ]


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