Answer to Question #268342 in Statistics and Probability for A@i

Question

The diameter of an electric cable, Say x is assumed to be Continuous random a Variable with Pdf f(x) = 6x * (1 - x), 0 <= x <= 1 check that f(x) is a P.d.f. ,Determine a&b such that P(x<b) = P(x>b) ,Find P(0.1<x <0.3)

Answer 1

a)

$$\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{0}^{1}6x(1-x)dx$$

$$=[3x^2-2x^3]\begin{matrix} 1 \\ 0 \end{matrix}=1$$

b)

$$P(X<b)=\displaystyle\int_{0}^{b}6x(1-x)dx$$

$$=[3x^2-2x^3]\begin{matrix} b \\ 0 \end{matrix}=3b^2-2b^3, 0<b<1$$

$$P(x>b)=1-P(X\leq b)$$

$$=1-(3b^2-2b^3)$$

$$P(x<b)=P(x>b)$$

$$3b^2-2b^3=1-(3b^2-2b^3)$$

$$4b^3-6b^2+1=0$$

$$4b^3-2b^2-(4b^2-1)=0$$

$$2b^2(2b-1)-(2b-1)(2b+1)=0$$

$$(2b-1)(2b^2-2b-1)=0$$

$$2b-1=0=>b_1=1/2$$

$$2b^2-2b-1=0$$

$$D=(-2)^2-4(2)(-1)=12$$

$$b=\dfrac{2\pm\sqrt{12}}{2(2)}=\dfrac{1\pm\sqrt{3}}{2}$$

$$b_2=\dfrac{1-\sqrt{3}}{2}<0, b_3=\dfrac{1+\sqrt{3}}{2}>1$$

Therefore $b=1/2$

c)

$$P(0.1<X<0.3)=\displaystyle\int_{0.1}^{0.3}6x(1-x)dx$$

$$=[3x^2-2x^3]\begin{matrix} 0.3 \\ 0.1 \end{matrix}$$

$$=3(0.3)^2-2(0.3)^3-(3(0.1)^2-2(0.1)^3)$$

$$=0.188$$