Exercise Solution 3.29

Exercise Solution 3.29

  1. By [3.94], M ~ N(8, 36).
  2. By [3.105] and [3.106], L ~ .
  3. By [3.101] and [3.102], E ~ Λ(e5e16 – e10).
  4. We cannot say. We are not told if N1 and N2 are joint-normal.
  5. Because N1 and N2 are independent, they are joint-normal. Hence, M is normal. By [3.27] and [3.28], M ~ N(2, 6).
  6. H ~ χ2(1, 0)
  7. H ~ χ2(2, 25)
  8. This is a nonstandard distribution. It is clearly not normal because C is strictly positive.

 
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