Find the standard​ deviation, s, of sample data summarized in the frequency distribution table below by using the formula​ below, where 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.
sequalsStartRoot StartFraction n left bracket Summation from nothing to nothing left parenthesis f times x squared right parenthesis right bracket minus left bracket Summation from nothing to nothing left parenthesis f times x right parenthesis right bracket squared Over n left parenthesis n minus 1 right parenthesis EndFraction EndRoot
Interval
20​-29
30​-39
40​-49
50​-59
60​-69
70​-79
80​-89
 
Frequency
2
1
6
3
9
34
35


To calculate the standard deviation \( s \) of the sample data using the provided frequency distribution, we'll follow the steps described in the formula:

\[
s = \sqrt{\frac{n \sum (f \cdot x^2) - \left(\sum (f \cdot x)\right)^2}{n(n-1)}}
\]

### Step-by-Step Calculation

1. **Calculate the midpoint \( x \) for each class interval:**
   The midpoint for each class interval is found by averaging the lower and upper bounds of the interval.

\[
\text{Midpoint} = \frac{\text{Lower Bound} + \text{Upper Bound}}{2}
\]

For each class interval:
- 20–29: \( \frac{20 + 29}{2} = 24.5 \)
- 30–39: \( \frac{30 + 39}{2} = 34.5 \)
- 40–49: \( \frac{40 + 49}{2} = 44.5 \)
- 50–59: \( \frac{50 + 59}{2} = 54.5 \)
- 60–69: \( \frac{60 + 69}{2} = 64.5 \)
- 70–79: \( \frac{70 + 79}{2} = 74.5 \)
- 80–89: \( \frac{80 + 89}{2} = 84.5 \)

2. **Create a table with midpoints \( x \), frequencies \( f \), and calculate \( f \cdot x \) and \( f \cdot x^2 \) for each class interval:**

\[
\begin{array}{|c|c|c|c|c|}
\hline
\text{Class Interval} & \text{Frequency } f & \text{Midpoint } x & f \cdot x & f \cdot x^2 \\
\hline
20–29 & 2 & 24.5 & 2 \cdot 24.5 = 49 & 2 \cdot 24.5^2 = 2 \cdot 600.25 = 1200.5 \\
30–39 & 1 & 34.5 & 1 \cdot 34.5 = 34.5 & 1 \cdot 34.5^2 = 1 \cdot 1190.25 = 1190.25 \\
40–49 & 6 & 44.5 & 6 \cdot 44.5 = 267 & 6 \cdot 44.5^2 = 6 \cdot 1980.25 = 11881.5 \\
50–59 & 3 & 54.5 & 3 \cdot 54.5 = 163.5 & 3 \cdot 54.5^2 = 3 \cdot 2967.25 = 8901.75 \\
60–69 & 9 & 64.5 & 9 \cdot 64.5 = 580.5 & 9 \cdot 64.5^2 = 9 \cdot 4160.25 = 37442.25 \\
70–79 & 34 & 74.5 & 34 \cdot 74.5 = 2533 & 34 \cdot 74.5^2 = 34 \cdot 5550.25 = 188708.5 \\
80–89 & 35 & 84.5 & 35 \cdot 84.5 = 2957.5 & 35 \cdot 84.5^2 = 35 \cdot 7140.25 = 249908.75 \\
\hline
\end{array}
\]

3. **Compute the sums \( \sum (f \cdot x) \) and \( \sum (f \cdot x^2) \):**
   
- \( \sum (f \cdot x) = 49 + 34.5 + 267 + 163.5 + 580.5 + 2533 + 2957.5 = 6585 \)
- \( \sum (f \cdot x^2) = 1200.5 + 1190.25 + 11881.5 + 8901.75 + 37442.25 + 188708.5 + 249908.75 = 498233.5 \)

4. **Calculate \( n \), the total number of sample values:**

\[
n = 2 + 1 + 6 + 3 + 9 + 34 + 35 = 90
\]

5. **Apply the standard deviation formula:**

\[
s = \sqrt{\frac{90 \times 498233.5 - (6585)^2}{90 \times 89}}
\]

Let's calculate the expression inside the square root:

\[
90 \times 498233.5 = 44841015
\]
\[
(6585)^2 = 43372025
\]
\[
44841015 - 43372025 = 1468990
\]
\[
90 \times 89 = 8010
\]
\[
s = \sqrt{\frac{1468990}{8010}} = \sqrt{183.43} = 13.54
\]

### Final Answer:
The computed standard deviation \( s \) is approximately 13.54.

### Comparison:
The given standard deviation from the original list of data values is 11.1. The computed standard deviation (13.54) is slightly higher than the original standard deviation (11.1).

Would you like any additional details or further explanation? 

Here are five related questions you might find helpful:
1. How do you calculate the variance from the standard deviation?
2. What is the difference between population and sample standard deviation?
3. Why is the standard deviation important in statistical analysis?
4. How does changing class intervals affect the standard deviation?
5. What are the limitations of using grouped data to compute standard deviation?

**Tip**: The standard deviation is sensitive to outliers—extreme values can significantly affect it!

Find the standard​ deviation, s, of sample data summarized in the frequency distribution table below by using the formula​ below, where 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.

Learn how to calculate the standard deviation of sample data summarized in a frequency distribution table using the formula for grouped data. This example involves computing midpoints, frequencies, and comparing the result to an original standard deviation of 11.1.

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