Suppose that the random variable X is defined as X=0.6X1+0.3X2+0.1X3, where the probability distributions of the independent random variables X1,X2,andX3 are fX1(x)=(5x)0.1x×0.95−x fX2(x)=(5x)0.4x×0.65−x fX3(3)=(5x)0.6x×0.45−x x=0,1,2,3,4,5 a) Evaluate the mean of X. b) Evaluate the variance of X.

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Suppose that the random variable X is defined as X=0.6X1+0.3X2+0.1X3, where the probability distributions of the independent random variables X1,X2,andX3 are  fX1(x)=(5x)0.1x×0.95−x  fX2(x)=(5x)0.4x×0.65−x  fX3(3)=(5x)0.6x×0.45−x  x=0,1,2,3,4,5  a) Evaluate the mean of X.  b) Evaluate the variance of X.

Archie Griffin · Accepted Answer

Step 1  Given:  X=0.6X1+0.3X2+0.1X3  Where all three are independent  fX1(x)=(5x)0.1x×0.95−x  fX2(x)=(5x)0.4x×0.65−x  fX3(3)=(5x)0.6x×0.45−x  x=0,1,2,3,4,5  Step 2  It can be seen that all the three distributions follows binomial distribution  So, Mean of binomial is  n×p  Variance of binomial is  n×p×q  So, X1∼B∈(5,0.1)  E[X1]=5×0.1=0.5  Var[X1]=5×0.1×0.9=0.45  X2∼B∈(5,0.4)  E[X2]=5×0.4=2  Var[X2]=5×0.4×0.6=1.2  X3∼B∈(5,0.6)  E[X3]=5×0.6=3  Var[X3]=5×0.6×0.4=1.2  Step 3  a) Mean of X  E[X]=0.6×E[X1]+0.3×E[X2]+0.1×E[X3]  E[X]=0.6×0.5+0.3×2+0.1×3  E[X]=1.2  Hence, the mean is 1.2  Step 4  b) If the events are independent, then the variance is  Var[A+B]=Var[A]+Var[B]  Here, all are independent, so the variance of X can be calculated as follows  Var[X]=0.62×Var[X1]+0.32×Var[X2]+0.12×Var[X3]  Var[X]=0.62×0.45+0.32×1.2+0.12×1.2

Suppose that the random variable X is defined as X=0.6X_{1}+0.3X_{2}+0.1X_{3},

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