f X~ N(0, 1), Y~N(0, 4), Z~N(0, 25). X, Y, Z are pairwise independent. Then X+Y+Z ~ N(0, 64)X+Y+Z ~ N(0, 29)X+Y+Z ~N(0, 30)X+Y+Z~ N(0,50)
Question
f X~ N(0, 1), Y~N(0, 4), Z~N(0, 25). X, Y, Z are pairwise independent. Then X+Y+Z ~ N(0, 64)X+Y+Z ~ N(0, 29)X+Y+Z ~N(0, 30)X+Y+Z~ N(0,50)
Solution
The sum of independent normal random variables is also a normal random variable. The mean of the sum is the sum of the means, and the variance of the sum is the sum of the variances.
Given that X, Y, and Z are independent normal random variables with means 0 and variances 1, 4, and 25 respectively, the sum X+Y+Z is a normal random variable with mean 0+0+0=0 and variance 1+4+25=30.
Therefore, X+Y+Z ~ N(0, 30).
Similar Questions
If X~ N(0, 1), Y~N(0, 4), Z~N(0, 25). X, Y, Z are pairwise independent. Then
f X~ N(0, 1), Y~N(0, 4), Z~N(0, 25). X, Y, Z are pairwise independent. Then X+Y+Z ~ N(0, 64)X+Y+Z ~ N(0, 29)X+Y+Z ~N(0, 30)X+Y+Z~ N(0,50)
f (x, y, z) = x2 y3 z4 .
Consider the Boolean function of four variables: f(w,x,y,z) = ∑(1,3,4,6,9,11,12,14) The function is: (A) Independent of one variables(B) Independent of two variables(C) Independent of three variables(D) Dependent on all the variables
Find the Conjunctive normal for and Disjunctive normal form ofgiven Boolean function f(x, y, z)=(xvy) (xvz).
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