Find the expected value of the random variable X and its variance having the following density function:
f(x) = 5 (1-x4) 0<x<2
= 0 elsewhere
Full details math and explain.
Expected value
E(X)=5∫02x(1−x4)dx=5∫02x−x5dx=5[x22−x66]02=5[(222−266)−0]=−1303E(X)=5∫_0^2x(1-x^4 )dx=5∫_0^2x-x^5 dx=5[\frac{x^2}{2}-\frac{x^6}{6}]_0^2=5[(\frac{2^2}{2}-\frac{2^6}{6})-0]=-\frac{130}{3}E(X)=5∫02x(1−x4)dx=5∫02x−x5dx=5[2x2−6x6]02=5[(222−626)−0]=−3130
Variance
Var(X)=E(X2)−(E(X))2Var(X)=E(X^2)-(E(X))^2Var(X)=E(X2)−(E(X))2
E(X2)=5∫02x2(1−x4)dx=5∫02x2−x6dx=5[x33−x77]02=5[(233−277)−0]=−164021E(X^2)=5∫_0^2x^2 (1-x^4 )dx=5∫_0^2x^2-x^6 dx=5[\frac{x^3}{3}-\frac{x^7}{7}]_0^2=5[(\frac{2^3}{3}-\frac{2^7}{7})-0]=-\frac{1640}{21}E(X2)=5∫02x2(1−x4)dx=5∫02x2−x6dx=5[3x3−7x7]02=5[(323−727)−0]=−211640
Var(X)=−164021−(−1303)2=−12322063=−1955.8730Var(X)=-\frac{1640}{21}-(-\frac{130}{3})^2=-\frac{123220}{63}=-1955.8730Var(X)=−211640−(−3130)2=−63123220=−1955.8730
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