Part 1 of 2 Points: 0 of 3 Save 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 $n$ 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. \[ \begin{array}{l} s=\sqrt{\frac{n\left[\sum\left(f \cdot x^{2}\right)\right]-\left[\sum(f \cdot x)\right]^{2}}{n(n-1)}} \\ \begin{array}{c|c|c|c|c|c|c|c|c} \text { Interval } & 30-36 & 37-43 & 44-50 & 51-57 & 58-64 & 65-71 & 72-78 \\ \hline \text { Frequency } & 1 & 1 & 3 & 4 & 18 & 34 \end{array} \\ \end{array} \] 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 \(33.0, 40.0, 47.0, 54.0, 61.0, 68.0, 75.0\).

Step 2 ：Next, calculate the sum of the product of the frequency and the square of the class midpoint, which is \(245174.0\), and the sum of the product of the frequency and the class midpoint, which is \(3840.0\).

Step 3 ：Also, calculate the total number of sample values, which is the sum of the frequencies. The total number of sample values is \(61\).

Step 4 ：Substitute these values into the formula for the standard deviation: \(s=\sqrt{\frac{n[\sum(f \cdot x^{2})]-[\sum(f \cdot x)]^{2}}{n(n-1)}}\).

Step 5 ：After calculating, the standard deviation is approximately \(7.575016456545813\).

Step 6 ：Finally, round the standard deviation to one decimal place. The standard deviation of the sample data summarized in the frequency distribution table is approximately \(\boxed{7.6}\). This is less than the standard deviation obtained from the original list of data values, which is \(11.1\).