[tex]f_X(x)=\begin{cases}\dfrac3{x^4}&\text{for }x>1\\\\0&\text{otherwise}\end{cases}[/tex]
This distribution has expectation
[tex]E[X]=\displaystyle\int_{-\infty}^\infty xf_X(x)\,\mathrm dx=\int_1^\infty\frac3{x^3}\,\mathrm dx=\frac32[/tex]
a. The probability that [tex]X[/tex] falls below the average/expectation is
[tex]P\left(X<\dfrac32\right)=\displaystyle\int_{-\infty}^{3/2}f_X(x)\,\mathrm dx=\int_1^{3/2}\frac3{x^4}\,\mathrm dx=\boxed{\frac{19}{27}}[/tex]
b. Denote by [tex]X_{(3)}[/tex] the largest of the three claims [tex]X_1,X_2,X_3[/tex]. Then the density of this maximum order statistic is
[tex]f_{X_{(3)}}(x)=3f_X(x)F_X(x)^2[/tex]
where [tex]F_X(x)=P(X\le x)[/tex] is the distribution function for [tex]X[/tex]. This is given by
[tex]F_X(x)=\displaystyle\int_{-\infty}^xf_X(t)\,\mathrm dt=\begin{cases}0&\text{for }x<1\\\\\displaystyle\int_1^xf_X(t)\,\mathrm dt=1-\frac1{x^3}&\text{for }x\ge1\end{cases}[/tex]
So we have
[tex]f_{X_{(3)}}(x)=\begin{cases}0&\text{for }x<1\\\dfrac9{x^4}\left(1-\dfrac1{x^3}\right)^2&\text{for }x>1\end{cases}[/tex]
and the expectation is
[tex]E[X_{(3)}]=\displaystyle\int_{-\infty}^\infty xf_{X_{(3)}}(x)\,\mathrm dx=\int_1^\infty\frac9{x^3}\left(1-\frac1{x^3}\right)^2\,\mathrm dx=\frac{81}{40}=\boxed{2.025}[/tex]
c. Denote by [tex]X_{(1)}[/tex] the smallest of the three claims. [tex]X_{(1)}[/tex] has density
[tex]f_{X_{(1)}}(x)=3f_X(x)(1-F_X(x))^2=\begin{cases}0&\text{for }x<1\\\\\dfrac9{x^{10}}&\text{for }x>1\end{cases}[/tex]
so the expectation is
[tex]E[X_{(1)}]=\displaystyle\int_{-\infty}^\infty xf_{X_{(1)}}(x)\,\mathrm dx=\int_1^\infty\frac9{x^9}\,\mathrm dx=\frac98=\boxed{1.125}[/tex]
