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​-26
27​-33
34​-40
41​-47
48​-54
55​-61
62​-68
 
Frequency
2
3
5
1
15
34
33

To find the sample standard deviation \( s \) from the frequency distribution using the given formula, we need to follow these steps:

1. **Identify class midpoints**: These are the midpoints of each class interval.
2. **Multiply each class midpoint by the corresponding frequency**: This gives \( f \times x \).
3. **Multiply each class midpoint squared by the corresponding frequency**: This gives \( f \times x^2 \).
4. **Calculate the sum of \( f \times x \) and the sum of \( f \times x^2 \)**.
5. **Compute the total number of sample values \( n \)**: This is the sum of the frequencies.
6. **Apply the formula for the standard deviation**.

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

Let's organize the steps:

### Step 1: Class Midpoints

The class midpoints \( x \) are calculated as the average of the lower and upper boundaries of each class interval.

\[
x = \frac{\text{lower boundary} + \text{upper boundary}}{2}
\]

So, for each interval:
- \( 20-26 \): \( x = \frac{20 + 26}{2} = 23 \)
- \( 27-33 \): \( x = \frac{27 + 33}{2} = 30 \)
- \( 34-40 \): \( x = \frac{34 + 40}{2} = 37 \)
- \( 41-47 \): \( x = \frac{41 + 47}{2} = 44 \)
- \( 48-54 \): \( x = \frac{48 + 54}{2} = 51 \)
- \( 55-61 \): \( x = \frac{55 + 61}{2} = 58 \)
- \( 62-68 \): \( x = \frac{62 + 68}{2} = 65 \)

### Step 2: Frequency Table

| Interval | Frequency \( f \) | Midpoint \( x \) | \( f \cdot x \) | \( f \cdot x^2 \) |
|----------|------------------|-----------------|----------------|-----------------|
| 20-26    | 2                | 23              | \( 2 \cdot 23 = 46 \) | \( 2 \cdot 23^2 = 1058 \) |
| 27-33    | 3                | 30              | \( 3 \cdot 30 = 90 \) | \( 3 \cdot 30^2 = 2700 \) |
| 34-40    | 5                | 37              | \( 5 \cdot 37 = 185 \) | \( 5 \cdot 37^2 = 6845 \) |
| 41-47    | 1                | 44              | \( 1 \cdot 44 = 44 \) | \( 1 \cdot 44^2 = 1936 \) |
| 48-54    | 15               | 51              | \( 15 \cdot 51 = 765 \) | \( 15 \cdot 51^2 = 39015 \) |
| 55-61    | 34               | 58              | \( 34 \cdot 58 = 1972 \) | \( 34 \cdot 58^2 = 114328 \) |
| 62-68    | 33               | 65              | \( 33 \cdot 65 = 2145 \) | \( 33 \cdot 65^2 = 139425 \) |

### Step 3: Summations

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 from a frequency distribution table using midpoints and the formula s = sqrt{ (n * Σ(f * x^2) - (Σ(f * x))^2) / (n * (n - 1)) }. This detailed step-by-step solution covers important statistical concepts like frequency distribution and standard deviation, suitable for students in Grades 10-12.

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