Find the standard deviation, s, of sample data summarized in the frequency distribution table given below by using the formula below, where x represents the class nidpoint, 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, 9.0. n(n – 1) Interval 20-26 27-33 34-40 41-47 48-54 55-61 Frequency 4 24 40 19 7 Standard deviation = (Round to one decimal place as needed.) Consider a difference of 20% between two values of a standard deviation to be significant. How does this computed value compare with the given standard deviation 9.0? O A. The computed value is significantly less than the given value. O B. The computed value is significantly greater than the given value. O C. The computed value is not significantly different from the given value.

MATLAB: An Introduction with Applications
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**Title: Calculating Standard Deviation from a Frequency Distribution Table**

In this exercise, we aim to find the standard deviation, \(s\), of a dataset summarized in the frequency distribution table below. The standard deviation will be calculated using the formula given, where \(x\) represents the class midpoint, \(f\) represents the class frequency, and \(n\) represents the total number of sample values. Additionally, we will compare the computed standard deviation to the standard deviation obtained from the original list of data values, which is 9.0.

The formula for the standard deviation in this context is:

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

**Frequency Distribution Table:**

| Interval | Frequency |
|----------|-----------|
| 20-26    | 4         |
| 27-33    | 24        |
| 34-40    | 40        |
| 41-47    | 19        |
| 48-54    | 7         |
| 55-61    | 5         |

**Step-by-Step Calculation:**

1. **Determine the midpoints (x) for each class interval**:
   \[ \text{Midpoint} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \]
   
2. **Multiply each midpoint by its class frequency to find \( f \cdot x \)** for each class interval and sum the results.

3. **Multiply the square of each midpoint by its class frequency to find \( f \cdot x^2 \)** for each class interval and sum the results.

4. **Substitute these sums into the standard deviation formula** and solve for \(s\).

**Result Interpretation:**

After computing the standard deviation, we need to compare the computed value to the given value of 9.0. Consider a difference of 20% between two values of standard deviation to be significant.

**Comparison Options:**
- \( \circ \) **A. The computed value is significantly less than the given value.**
- \( \circ \) **B. The computed value is significantly greater than the given value.**
- \( \circ \) **C. The computed value
Transcribed Image Text:--- **Title: Calculating Standard Deviation from a Frequency Distribution Table** In this exercise, we aim to find the standard deviation, \(s\), of a dataset summarized in the frequency distribution table below. The standard deviation will be calculated using the formula given, where \(x\) represents the class midpoint, \(f\) represents the class frequency, and \(n\) represents the total number of sample values. Additionally, we will compare the computed standard deviation to the standard deviation obtained from the original list of data values, which is 9.0. The formula for the standard deviation in this context is: \[ s = \sqrt{\frac{n \left( \sum (f \cdot x^2) \right) - \left( \sum (f \cdot x) \right)^2}{n(n - 1)}} \] **Frequency Distribution Table:** | Interval | Frequency | |----------|-----------| | 20-26 | 4 | | 27-33 | 24 | | 34-40 | 40 | | 41-47 | 19 | | 48-54 | 7 | | 55-61 | 5 | **Step-by-Step Calculation:** 1. **Determine the midpoints (x) for each class interval**: \[ \text{Midpoint} = \frac{\text{Lower Limit} + \text{Upper Limit}}{2} \] 2. **Multiply each midpoint by its class frequency to find \( f \cdot x \)** for each class interval and sum the results. 3. **Multiply the square of each midpoint by its class frequency to find \( f \cdot x^2 \)** for each class interval and sum the results. 4. **Substitute these sums into the standard deviation formula** and solve for \(s\). **Result Interpretation:** After computing the standard deviation, we need to compare the computed value to the given value of 9.0. Consider a difference of 20% between two values of standard deviation to be significant. **Comparison Options:** - \( \circ \) **A. The computed value is significantly less than the given value.** - \( \circ \) **B. The computed value is significantly greater than the given value.** - \( \circ \) **C. The computed value
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