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
1
3
1
4
9
33
37
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Part 1
Standard deviationequals
  
enter your response here ​(Round to one decimal place as​ needed.) me regales el standard deviationequals por favro

To compute the standard deviation \( s \) for the given frequency distribution, we use the following formula:

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

Where:
- \( f \) is the class frequency,
- \( x \) is the class midpoint,
- \( n \) is the total number of sample values.

### Step 1: Calculate the midpoints for each class interval.
The midpoint \( x \) of a class interval is calculated as:

\[
x = \frac{{\text{Lower bound} + \text{Upper bound}}}{2}
\]

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

### Step 2: Create a table to compute \( fx \), \( fx^2 \), and the total number of samples \( n \).

| Interval | Frequency (\( f \)) | Midpoint (\( x \)) | \( fx \)       | \( fx^2 \)       |
|----------|---------------------|-------------------|----------------|------------------|
| 20-29    | 1                   | 24.5              | 1 * 24.5 = 24.5 | 1 * (24.5)^2 = 600.25 |
| 30-39    | 3                   | 34.5              | 3 * 34.5 = 103.5 | 3 * (34.5)^2 = 3570.75 |
| 40-49    | 1                   | 44.5              | 1 * 44.5 = 44.5 | 1 * (44.5)^2 = 1980.25 |
| 50-59    | 4                   | 54.5              | 4 * 54.5 = 218 | 4 * (54.5)^2 = 11884.5 |
| 60-69    | 9                   | 64.5              | 9 * 64.5 = 580.5 | 9 * (64.5)^2 = 37461.75 |
| 70-79    | 33                  | 74.5              | 33 * 74.5 = 2458.5 | 33 * (74.5)^2 = 184216.125 |
| 80-89    | 37                  | 84.5              | 37 * 84.5 = 3126.5 | 37 * (84.5)^2 = 265168.625 |

### Step 3: Compute the sums.
- \( \sum(fx) = 24.5 + 103.5 + 44.5 + 218 + 580.5 + 2458.5 + 3126.5 = 6556 \)
- \( \sum(fx^2) = 600.25 + 3570.75 + 1980.25 + 11884.5 + 37461.75 + 184216.125 + 265168.625 = 504882.25 \)
- Total number of samples \( n = 1 + 3 + 1 + 4 + 9 + 33 + 37 = 88 \)

### Step 4: Apply the formula.

\[
s = \sqrt{ \frac{ 88 \times 504882.25 - (6556)^2 }{88 \times 87} }
\]

First, calculate the numerator:
\[
88 \times 504882.25 = 44429638
\]
\[
(6556)^2 = 42981376
\]
\[
\text{Numerator} = 44429638 - 42981376 = 1448258
\]

Now, calculate the denominator:
\[
88 \times 87 = 7656
\]

Finally, calculate the standard deviation:
\[
s = \sqrt{ \frac{1448258}{7656} } = \sqrt{189.11} = 13.75
\]

### Step 5: Round to one decimal place.
The standard deviation is approximately \( s = 13.8 \).

### Comparison:
The computed standard deviation \( 13.8 \) is larger than the given standard deviation from the original data, \( 11.1 \).

---

Would you like further clarification on the steps? 

Here are 5 related questions for deeper understanding:
1. What is the difference between population and sample standard deviation?
2. How do you interpret the midpoint in a frequency distribution?
3. Can the formula for variance be used in place of standard deviation, and how would that look?
4. Why is the computed standard deviation different from the original data’s standard deviation?
5. How does changing the number of samples affect the standard deviation?

**Tip**: Always round to the nearest decimal place as required by your context to ensure the best possible accuracy for your answers.

Find the standard deviation, s, of sample data summarized in the frequency distribution table below by using the formula, 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 from a frequency distribution table using the midpoint and frequency. This example involves intervals 20-29 to 80-89 with varying frequencies and demonstrates the step-by-step calculation. The final standard deviation is compared to the original value of 11.1.

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