# Exercise Solution 3.29

- By [3.94],
*M ~ N*(8, 36). - By [3.105] and [3.106],
*L*~ . - By [3.101] and [3.102],
*E*~ Λ(*e*^{5},*e*^{16}–*e*^{10}). - We cannot say. We are not told if
*N*_{1}and*N*_{2}are joint-normal. - Because
*N*_{1}and*N*_{2}are independent, they are joint-normal. Hence,*M*is normal. By [3.27] and [3.28],*M*~*N*(2, 6). *H*~ χ^{2}(1, 0)*H*~ χ^{2}(2, 25)- This is a nonstandard distribution. It is clearly not normal because
*C*is strictly positive.