Respuesta :
Answer:
2025
Step-by-step explanation:
Let X1, X2, X3 mean the three cases, and Y = max(X1, X2, X3)
F(y)=P(max(X1,X2,X3)≤y)=P(X1 ≤y) P(X2 ≤y) P(X3 ≤y) =(Z y
1 3x^-4dx)^3= (1-1/y^3)^3
=9(1/2-2/5+1/8)
=81/40
=2.025
Since the units are a huge number of dollars, the appropriate response is 2025 .
Answer:
The expected value of the largest of the three claims is $ 2025.00
Step-by-step explanation:
To solve the question, if we have the three claims given by
C₁, C₂, C₃ and the let the maximum claim be denoted by M = max(C₁, C₂, C₃)
If the expected value of M is expressed as E(M) and the common density function is F(m) with p,d,f f(m) = F'(m) of M
Therefore, F(m) = P (max C₁, C₂, C₃) ≤ M) = P (C₁ ≤m) × P (C₂ ≤m) × P (C₃ ≤m)
[tex](\int\limits^y_1 {3x^{-4}} \, dx )^3[/tex]= [tex](1-\frac{1}{m^3} )^{3[/tex]
From which
F'(m) = [tex]\frac{3}{m^4}\times 3\times (1-\frac{1}{m^3} )^{2[/tex] = [tex]\frac{9}{m^4} (1-\frac{1}{m^3} )^{2[/tex]
E(M) = [tex]\int\limits^\infty_1 {m f(m)} \, dm = \int\limits^\infty_1 {\frac{9}{m^3} (1-\frac{1}{m^3} )^{2\, dm[/tex]
Which gives [tex]9\times \int\limits^\infty_1 {\frac{1}{m^3} (1-\frac{1}{m^3} )^{2\, dm[/tex]
= [tex]9\times \int\limits^\infty_1 { (\frac{1}{m^3} -\frac{2}{m^6} + \frac{1}{m^9} )\, dm[/tex]
= 9×[tex](\frac{1}{2}-\frac{2}{5}+\frac{1}{8})[/tex] = $ 2.025 × 10³
=$ 2025.00
