Answer:
Step-by-step explanation:
A fluctuating electric current II may be considered a uniformly distributed random variable over the interval (9, 11)
[tex]p_1=\{^{\frac{1}{2}:9\leq i\leq 11}_{0:otherwise[/tex]
Now define
[tex]p = 2I^2[/tex]
[tex]\Rightarrow I^2=(\frac{p}{2} )\\\\\Rightarrow I=(\frac{p}{2} )^{\frac{1}{2} }\\\\\Rightarrow h^{-1}(p)=(\frac{p}{2} )^{\frac{1}{2}}[/tex]
[tex]\frac{dh^{-1}}{dp} =\frac{d[h^{-1}(p)]}{dp} \\\\=\frac{d(p/2)^{\frac{1}{2} }}{dp}[/tex]
[tex]=\frac{1}{2} \times \frac{1}{2} (\frac{p}{2} )^{{\frac{1}{2}-1} }\\\\=\frac{1}{4}(\frac{p}{2} )^{{\frac{1}{2}-1} }\\\\=\frac{1}{2}(\frac{2}{p} )^{{\frac{1}{2}} }[/tex]
using the transformation method, we get
[tex]f_p(p)=f_1(h^{-1}(p))|\frac{d[h^{-1}(p)]}{dp} |\\\\=\frac{1}{2} \times \frac{1}{4} (\frac{2}{p} )^{\frac{1}{2} }\\\\=\frac{1}{8} (\frac{2}{p} )^{\frac{1}{2} }[/tex]
[tex]f_p(p)=\{^{\frac{1}{8} (\frac{2}{p} )^{\frac{1}{2} },162\leqp\leq 242} }_{0,otherwise}[/tex]
