Answer
$P(30-39)=\frac{3807}{6427}\approx0.592$
If we randomly select 1000 mothers involved in a multiple birth we would expect about 592 to be mother 30 to 39 years old.
Work Step by Step
N(S) = 6427, N(30-34) = 2262 and N(35-39) = 1545.
$P(30-34)=\frac{N(30-34)}{N(S)}=\frac{2262}{6427}$
$P(35-39)=\frac{N(35-39)}{N(S)}=\frac{1545}{6427}$
The events "mother 30 to 34 years old involved in a multiple birth" and "mother 35 to 39 years old involved in a multiple birth" are mutually exclusive. So:
$P(30-39) = P(30-34$ or $35-39)$
$P(30-39) = P(30-34) + P(35-39)=\frac{2262}{6427}+\frac{1545}{6427}=\frac{3807}{6427}\approx0.592$
