Question #268342

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)

1
Expert's answer
2021-11-21T13:34:12-0500

a)


f(x)dx=016x(1x)dx\displaystyle\int_{-\infin}^{\infin}f(x)dx=\displaystyle\int_{0}^{1}6x(1-x)dx

=[3x22x3]10=1=[3x^2-2x^3]\begin{matrix} 1 \\ 0 \end{matrix}=1

b)


P(X<b)=0b6x(1x)dxP(X<b)=\displaystyle\int_{0}^{b}6x(1-x)dx

=[3x22x3]b0=3b22b3,0<b<1=[3x^2-2x^3]\begin{matrix} b \\ 0 \end{matrix}=3b^2-2b^3, 0<b<1

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

=1(3b22b3)=1-(3b^2-2b^3)

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

3b22b3=1(3b22b3)3b^2-2b^3=1-(3b^2-2b^3)

4b36b2+1=04b^3-6b^2+1=0

4b32b2(4b21)=04b^3-2b^2-(4b^2-1)=0

2b2(2b1)(2b1)(2b+1)=02b^2(2b-1)-(2b-1)(2b+1)=0

(2b1)(2b22b1)=0(2b-1)(2b^2-2b-1)=0

2b1=0=>b1=1/22b-1=0=>b_1=1/2

2b22b1=02b^2-2b-1=0

D=(2)24(2)(1)=12D=(-2)^2-4(2)(-1)=12

b=2±122(2)=1±32b=\dfrac{2\pm\sqrt{12}}{2(2)}=\dfrac{1\pm\sqrt{3}}{2}

b2=132<0,b3=1+32>1b_2=\dfrac{1-\sqrt{3}}{2}<0, b_3=\dfrac{1+\sqrt{3}}{2}>1

Therefore b=1/2b=1/2


c)


P(0.1<X<0.3)=0.10.36x(1x)dxP(0.1<X<0.3)=\displaystyle\int_{0.1}^{0.3}6x(1-x)dx

=[3x22x3]0.30.1=[3x^2-2x^3]\begin{matrix} 0.3 \\ 0.1 \end{matrix}

=3(0.3)22(0.3)3(3(0.1)22(0.1)3)=3(0.3)^2-2(0.3)^3-(3(0.1)^2-2(0.1)^3)

=0.188=0.188


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS