Respuesta :
Answer:
a) 0.96
b) 0.016
c) 0.018
d) 0.982
e) x = 2
Step-by-step explanation:
We are given with the Probability density function f(x)= 2/x^3 where x > 1.
Firstly we will calculate the general probability that of P(a < X < b)
P(a < X < b) = [tex]\int_{a}^{b} \frac{2}{x^{3}} dx[/tex] = [tex]2\int_{a}^{b} x^{-3} dx[/tex]
= [tex]2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx[/tex] { Because [tex]\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a[/tex] }
= [tex]2[ \frac{x^{-2} }{-2}]^{b}_a[/tex] = [tex]\frac{2}{-2} [ x^{-2} ]^{b}_a[/tex]
= [tex]-1 [ b^{-2} - a^{-2} ][/tex] = [tex]\frac{1}{a^{2} } - \frac{1}{b^{2} }[/tex]
a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }
Comparing with general probability we get,
P(1 < X < 5) = [tex]\frac{1}{1^{2} } - \frac{1}{5^{2} }[/tex] = [tex]1 - \frac{1}{25}[/tex] = 0.96 .
b) P(X > 8) = P(8 < X < ∞) = 1/[tex]8^{2}[/tex] - 1/∞ = 1/64 - 0 = 0.016
c) P(6 < X < 10) = [tex]\frac{1}{6^{2} } - \frac{1}{10^{2} }[/tex] = [tex]\frac{1}{36} - \frac{1}{100 }[/tex] = 0.018 .
d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)
= [tex](\frac{1}{1^{2} } - \frac{1}{6^{2} })[/tex] + (1/[tex]10^{2}[/tex] - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982
e) We have to find x such that P(X < x) = 0.75 ;
⇒ P(1 < X < x) = 0.75
⇒ [tex]\frac{1}{1^{2} } - \frac{1}{x^{2} }[/tex] = 0.75
⇒ [tex]\frac{1} {x^{2} }[/tex] = 1 - 0.75 = 0.25
⇒ [tex]x^{2}[/tex] = [tex]\frac{1}{0.25}[/tex] ⇒ [tex]x^{2}[/tex] = 4 ⇒ x = [tex]2[/tex]
Therefore, value of x such that P(X < x) = 0.75 is 2.
Following are the calculation to the given points:
Given:
[tex]\to f(x) = \frac{2}{x^3} \ for\ x>1 \\\\[/tex]
For part a:
[tex]\to P(X<5) = \int^{5}_{1} \ (\frac{2}{x^3})\ dx = [ \frac{-1}{x^2}]^{5}_{1} = \frac{24}{25} = 0.96[/tex]
So, requiring probability[tex]=0.96[/tex]
For part b: [tex]\to P(x>8) = 1 - \int^{8}_{1} (\frac{2}{x^3})\ dx = 1 - [\frac{-1}{x^2}]^{8}_{1} = 1 – 0.98438 = 0.01562[/tex]
So, requiring probability[tex]= 0.01562[/tex]
For part c: [tex]\to P(6 < x < 10) = \int^{10}_{6} (\frac{2}{x^3}) \ dx = [-\frac{1}{x^2}]^{10}_{6} = 0.017778[/tex]
So, requiring probability[tex]= 0.017778[/tex]
For part d:
[tex]\to P(X<6 \ or\ x>10) = 1 -P(6< x < 10) = 1 -0.017778 = 0.982222[/tex]
So, requiring probability[tex]= 0.982222[/tex]
For question 3:
[tex]\to P(X<x) = 0.75 \\\\\to P(X<x) = \int^{x}_{1}(\frac{2}{x^3})\ dx = [-\frac{1}{x^2}]^{x}_{1} = 0.75\\\\ \to (-\frac{1}{x^2}) - (-\frac{1}{1^2}) = (-\frac{1}{x^2}) + 1 = 0.75 \\\\\to (-\frac{1}{x^2}) = 0.75 -1 = -0.25\\\\ \to (\frac{1}{x^2}) = 0.25\\\\ \to X^2 = \frac{1}{0.25} = 4 \\\\X^2= 2^2\\\\\to X =2[/tex]
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