Find the standard deviation, $s$, of sample data summarized in the frequency distribution table below by using the formula below, where $\mathrm{x}$ represents the class midpoint, $f$ represents the class frequency, and $\mathrm{n}$ represents the total number of sample values. Also, compare the computed standard deviation to the standard deviation obtained from the original list of data values, 11.1.
\[
s=\sqrt{\frac{n\left[\sum\left(f \cdot x^{2}\right)\right]-\left[\sum(f \cdot x)\right]^{2}}{n(n-1)}}
\]
\begin{tabular}{c|c|c|c|c|c|c|c|c}
Interval & $20-29$ & $30-39$ & $40-49$ & $50-59$ & $60-69$ & $70-79$ & $80-89$ \\
\hline Frequency & 1 & 2 & 4 & 2 & 12 & 37
\end{tabular}
Standard deviation $=$ (Round to one decimal place as needed.)

Step 1 ：First, calculate the class midpoints for each interval. The class midpoint is the average of the lower and upper bounds of each interval. The midpoints are [24.5, 34.5, 44.5, 54.5, 64.5, 74.5, 84.5].

Step 2 ：Next, calculate the sum of the product of the frequency and the square of the class midpoint, and the sum of the product of the frequency and the class midpoint. The sum of the product of the frequency and the square of the class midpoint is 272124.5, and the sum of the product of the frequency and the class midpoint is 3911.0.

Step 3 ：Calculate the total number of sample values, which is the sum of the frequencies. The total number of sample values is 58.

Step 4 ：Substitute these values into the formula to calculate the standard deviation. The standard deviation is \(\sqrt{\frac{58\left[272124.5\right]-\left[3911.0\right]^{2}}{58(58-1)}}\), which simplifies to 12.1.

Step 5 ：Final Answer: The standard deviation of the sample data summarized in the frequency distribution table is \(\boxed{12.1}\). This is slightly higher than the standard deviation obtained from the original list of data values, which is 11.1.