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A continuous random variable X has the p.d.f f(x) = kx2ex ; x 0 . Find k , mean & variance .
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For continuous Random Variable

f(x)dx=10f(x)dx=1[x0]0kx2exdx=1k[x2ex12xex(1)2+2ex(1)3]0=1k[2e0]=1k=12

E(X)=0=xf(x)dx

=0x[kx2ex]dx=k0x3ex]dx=k[x3ex13x2ex(1)2+6xex(1)36ex(1)4]0=12[6]=3

E(X2)=0x2f(x)dx=0x2[kx2ex]dx=k0x4ex]dx=k[x4ex14x3ex(1)2+12x2ex(1)324xex(1)4+24ex(1)5]0=12[24(1(1)5)]=12Var(X)=E(X2)E(X)2=129=3

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