sample values. Also, compare the computed standard deviation to the standard deviation obtained from the original list of data values, 11.1.
\[
\begin{array}{l}
s=\sqrt{\frac{n\left[\sum\left(f \cdot x^{2}\right)\right]-\left[\sum(f \cdot x)\right]^{2}}{n(n-1)}} \\
\begin{array}{c|c|c|c|c|c|c|c|}
\text { Interval } & 20-29 & 30-39 & 40-49 & 50-59 & 60-69 & 70-79 & 80-89 \\
\hline \text { Frequency } & 1 & 3 & 5 & 4 & 12 & 36 & 37
\end{array} \\
\end{array}
\]
Standard deviation $=$
(Round to one decimal place as needed.)

Step 1 ：Given the frequency distribution table, we first calculate the midpoint of each interval. The midpoint of an interval is calculated by adding the lower and upper bounds of the interval and dividing by 2.

Step 2 ：We then calculate the sum of the product of the frequency and the square of the midpoint, denoted as \(\sum\left(f \cdot x^{2}\right)\), and the sum of the product of the frequency and the midpoint, denoted as \(\sum(f \cdot x)\).

Step 3 ：Substitute these values into the formula for standard deviation of a frequency distribution: \[s=\sqrt{\frac{n\left[\sum\left(f \cdot x^{2}\right)\right]-\left[\sum(f \cdot x)\right]^{2}}{n(n-1)}}\]

Step 4 ：Given the values: intervals = [(20, 29), (30, 39), (40, 49), (50, 59), (60, 69), (70, 79), (80, 89)], frequencies = [1, 3, 5, 4, 12, 36, 37], midpoints = [24.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5], n = 98, sum_fx2 = 539874.5, sum_fx = 7151.0, we calculate the standard deviation to be approximately 13.648914523233302.

Step 5 ：Rounding to one decimal place, the final answer is \(\boxed{13.6}\).