Question

The diameter , Say X , of a electric cable , is assumed to be continuous random variable with p.d.f : f(x)= 6x(1-x) , 0<=x<=1.

(1) Check that the above is a P.D.F.

(2) Obtain an expression for the c. d. f. of X.

(3) Compute P(X<=1/2| 1/3<=X<=2/3), and

(4) Determine the value of k such that P(X<k)=P(X>k).

Answer 1

1) Every pdf must integrates to 1, so

$\int_{-\infty}^{+\infty} f(x) dx=\int_{0}^{1} 6x(1-x) dx=6\int_{0}^{1}x dx-6\int_{0}^{1}x^2 dx=3x^2|_0^1-2x^3|_0^1=3-2=1$

So, this function is valid pdf

2) F(x) - cdf, then $F(x)=\int_{-\infty}^{x} f(t) dt$ . So,

F(x) = 0 for $x<0$

$F(x)=\int_{0}^{x} 6t(1-t) dt=3t^2|_0^x-2t^3|_0^x=3x^2-2x^3$ for $0≤x<1$

F(x) = 1 for $x≥1$

3) $P(X≤0.5)=F(0.5)=3*0.5^2-2*0.5^3=0.5$

$P({\frac 1 3}\leq X≤{\frac 2 3})=F({\frac 2 3})-F({\frac 1 3})=3*({\frac 2 3})^2-2*({\frac 2 3})^3-3*({\frac 1 3})^2+2*({\frac 1 3})^3=\\={\frac 4 3}-{\frac {16} {27}}-{\frac 1 3}+{\frac 2 {27}}={\frac {13} {27}}$

4) $P(X<k)=P(X>k)\implies P(X<k)=0.5\implies k=0.5$ (already found in part 3).