Complete Question
The complete question is shown on the first uploaded image
Answer:
a  [tex]P(X <  5) =  0.960[/tex]
b [tex]P(X > Â 8) = 0.016[/tex]
c  [tex]P(6 < x < 10) =  0.018 [/tex]
d  [tex] P(X < 6 or  X > 10 ) =  0.982 [/tex]
e  [tex]X =  2[/tex]
Step-by-step explanation:
From the question we are told that
  The probability density function is  [tex]f(x) =  \frac{2}{x^3}[/tex] for  x > 1
Considering question a
 [tex]P(x < 5) = \int\limits^5_1 {\frac{2}{x^3} } \, dx[/tex]
=> [tex]P(X < Â 5) = Â [-\frac{1}{x^2} ]| Â \left \ 5} \atop {1}} \right.[/tex]
=>[tex]P(X < Â 5) = - \frac{1}{25} Â + Â \frac{1}{1^2}[/tex]
=> Â [tex]P(X < Â 5) = Â 0.960[/tex]
Considering question b
[tex]P(x > 8) =1 Â - \int\limits^6_1 {\frac{2}{x^3} } \, dx[/tex]
=> [tex]P(X > 8) =1- Â [-\frac{1}{x^2} ]| Â \left \ 8} \atop {1}} \right.[/tex]
=>[tex]P(X > Â 8) = 1 - [- \frac{1}{64} Â + Â \frac{1}{1^2}][/tex]
=>[tex]P(X > Â 8) = 0.016[/tex]
Considering question c
 [tex]P(6 < x < 10) = \int\limits^{10}_{6} {\frac{2}{x^3} } \, dx[/tex]
=> [tex]P(6 < x < 10) = Â [-\frac{1}{x^2} ]| Â \left \ 10} \atop {6}} \right.[/tex] Â
=>[tex]P(6 < x < 10) = Â [- \frac{1}{100} Â + Â \frac{1}{36}][/tex]
=>[tex]P(6 < x < 10) = Â 0.018 [/tex]
Considering question d
 [tex] P(X < 6 or  X > 10 ) = 1 - P(6 < x < 10) = 1 - \int\limits^{10}_{6} {\frac{2}{x^3} } \, dx[/tex]
=> [tex] P(X < 6 or  X > 10 ) =1-  [-\frac{1}{x^2} ]|  \left \ 10} \atop {6}} \right.[/tex]
=> [tex] P(X < 6 or  X > 10 ) =1- [- \frac{1}{100}  +  \frac{1}{36}][/tex] [/tex]
=> [tex] P(X < 6 or  X > 10 ) =  0.982 [/tex]
Considering question e
  [tex]P(X  <  x ) =  \int\limits^x_1 {\frac{2}{x^3} } \, dx  =  0.75[/tex]
  [tex]P(X  <  x ) =  [- \frac{1}{x^2} ]| \left \ x } \atop {1}} \right.  =  0.75[/tex]
  [tex]P(X  <  x ) =  - \frac{1}{x^2} - [- \frac{1}{1^2} ]= 0.75[/tex]
  [tex]P(X  <  x ) =  - \frac{1}{x^2} + 1 = 0.75[/tex]
   [tex]  - \frac{1}{x^2}  = -0.25[/tex]
   [tex]X =  2[/tex]
  Â
