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If 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

If 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)
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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).

This problem has been solved

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