here for this to be valid :
f(x) dx must be 1
f(x) dx = ax2 e-10x dx =a*(-x2e-10x/10-2xe-10x/100-2e-10x/1000) |0 =a/500 =1
a =500
for (d/dx)f(x) =(d/dx)*(500x2e-10x) =500*(2x*e-10x-10x2e-10x)
putting above equal to 0 ,
500*(2x*e-10x-10x2e-10x) =0
x =2/10 =0.2 =mode
therefore P(X>=mode )=P(X>=0.2) =1-P(X<=0.2)
=1- f(x) dx =1-500x2 e-10x dx =1-500*(-x2e-10x/10-2xe-10x/100-2e-10x/1000) |0.20 =1-0.3233
=0.6767
11. Let X be a continuous random variable with density function fare-102 for 10 f(1) =...
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