Question

for the continuous probability function f(X)=kx²e-x when X>0 find k​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

∫−∞∞f(x)dx=1∫∞0f(x)dx=1[x≥0]∫∞0kx2e−xdx=1k[x2e−x−1−2xe−x(−1)2+2e−x(−1)3]∞0=1k[−2e−0]=1∴k=12∫∞−∞f(x)dx=1∫0∞f(x)dx=1[x≥0]∫0∞kx2e−xdx=1k[x2e−x−1−2xe−x(−1)2+2e−x(−1)3]0∞=1k[−2e−0]=1∴k=12

E(X)=∫∞0=xf(x)dxE(X)=∫0∞=xf(x)dx

=∫∞0x[kx2e−x]dx=k∫∞0x3e−x]dx=k[x3e−x−1−3x2e−x(−1)2+6xe−x(−1)3−6e−x(−1)4]∞0=12[6]=3=∫0∞x[kx2e−x]dx=k∫0∞x3e−x]dx=k[x3e−x−1−3x2e−x(−1)2+6xe−x(−1)3−6e−x(−1)4]0∞=12[6]=3

E(X2)=∫∞0x2f(x)dx=∫∞0x2[kx2e−x]dx=k∫∞0x4e−x]dx=k[x4e−x−1−4x3e−x(−1)2+12x2e−x(−1)3−24xe−x(−1)4+24e−x(−1)5]∞0=12[−24(1(−1)5)]=12Var(X)=E(X2)−E(X)2=12−9=3E(X2)=∫0∞x2f(x)dx=∫0∞x2[kx2e−x]dx=k∫0∞x4e−x]dx=k[x4e−x−1−4x3e−x(−1)2+12x2e−x(−1)3−24xe−x(−1)4+24e−x(−1)5]0∞=12[−24(1(−1)5)]=12Var(X)=E(X2)−E(X)2=12−9=3