Part i
P (threads of the fitting is at least 0.5)
=P(X>0.5)=∫0.51f(x)dx=∫0.51(1+xB2xB−1)dx
But we can consider B as 6
∫0.51(1+x62x6−1)dxLet t=x3 then dt=3x2dx⟹3dt=x2dx⟹P(x>0.5)=∫0.5311+t22t3dt=31∫0.511+t22tdt=31[log(1+t2)]0.51=31log(916)
Part ii
=P(0.2<X<0.8)=∫0.20.8f(x)dx=∫0.20.8(1+xB2xB−1)dx
But we can consider B as 6
∫0.20.8(1+x62x6−1)dxLet t=x3 then dt=3x2dx⟹3dt=x2dx⟹P(0.2<x<0.8)=∫0.20.81+t22t3dt=31∫0.20.81+t22tdt=31[log(1+t2)]0.20.8=31log(23)
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