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(e⁵, e¹⁶ – 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.