The data on the right represent the number of live multiple-delivery births (three or more babies) in a particular year for women 15 to 54 years old. Use the data to complete parts (a) through (d) below. \begin{tabular}{|c|r|} \hline Age & $\begin{array}{r}\text { Number of Multiple } \\ \text { Births }\end{array}$ \\ \hline $15-19$ & 88 \\ \hline $20-24$ & 509 \\ \hline $25-29$ & 1632 \\ \hline $30-34$ & 2827 \\ \hline $35-39$ & 1849 \\ \hline $40-44$ & 380 \\ \hline $45-54$ & 118 \\ \hline \end{tabular} (a) Determine the probability that a randomly selected multiple birth for women 15-54 years old involved a mother 30 to 39 years old. $P(30$ to 39$)=$ (Type an integer or decimal rounded to three decimal places as needed.)

Step 1 ：First, we need to calculate the total number of multiple births. This is done by adding up all the numbers in the 'Number of Multiple Births' column. The total number of multiple births is \(88 + 509 + 1632 + 2827 + 1849 + 380 + 118 = 7403\).

Step 2 ：Next, we need to find the number of multiple births for women aged 30 to 39. This is done by adding up the numbers in the 'Number of Multiple Births' column for the age groups 30-34 and 35-39. The number of multiple births for women aged 30 to 39 is \(2827 + 1849 = 4676\).

Step 3 ：Finally, we can calculate the probability by dividing the number of multiple births for women aged 30 to 39 by the total number of multiple births. The probability is \(\frac{4676}{7403} = 0.6316358233148723\).

Step 4 ：Rounding to three decimal places, the final answer is \(\boxed{0.632}\).