Find the standard​ deviation, s, of the sample data summarized in the frequency distribution table below.  Since individual data values are​ unavailable, instead use class midpoints to represent the data values in each class. 
​Finally, compare your computed standard deviation to the standard deviation obtained from the original list of data​ values, which was 11.1.
Interval
20​-26
27​-33
34​-40
41​-47
48​-54
55​-61
62​-68
 
Frequency
1
1
2
1
11
38
32
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Part 1
Standard deviationequals
  
8.88 ​(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, 11.1​?
A.The computed value is significantly greater than the given value.
The computed value is significantly greater than the given value.
B.The computed value is significantly less than the given value.
The computed value is significantly less than the given value.
Your answer is correct.C.The computed value is not significantly different from the given value.
The computed value is not significantly different from the given value.

To find the standard deviation \( s \) of the sample data summarized in the frequency distribution table, we'll follow these steps:

### Steps:

1. **Determine class midpoints:**
   The midpoint of each class interval is calculated by averaging the lower and upper bounds of each class. 

2. **Use the midpoint and frequency:**
   Each midpoint represents the data value for all observations in that class. Multiply the midpoint by the corresponding frequency to compute the weighted mean.

3. **Calculate the mean:**
   The mean is found by dividing the sum of the midpoint-weighted frequencies by the total number of observations.

4. **Compute the variance:**
   For each class, subtract the mean from the midpoint, square the result, and multiply by the frequency. Then, sum these values and divide by \( n-1 \), where \( n \) is the total number of observations.

5. **Take the square root of the variance:**
   The square root of the variance gives the standard deviation \( s \).

### Frequency Distribution Table:

| Interval  | Midpoint \( x_i \) | Frequency \( f_i \) | 
|-----------|--------------------|---------------------|
| 20–26     | 23                 | 1                   |
| 27–33     | 30                 | 1                   |
| 34–40     | 37                 | 2                   |
| 41–47     | 44                 | 1                   |
| 48–54     | 51                 | 11                  |
| 55–61     | 58                 | 38                  |
| 62–68     | 65                 | 32                  |

### Calculation:

Using the midpoints \( x_i \), the frequencies \( f_i \), and the formula for sample standard deviation:

\[
s = \sqrt{\frac{\sum f_i (x_i - \bar{x})^2}{n-1}}
\]

where \( \bar{x} \) is the sample mean and \( n \) is the total number of observations.

### Given Result:

The computed standard deviation is 8.88, and the given standard deviation from the original data is 11.1.

### Comparison:

We compare the computed standard deviation (8.88) with the given standard deviation (11.1). To determine whether the difference is significant, we calculate the percentage difference:

\[
\text{Percentage difference} = \frac{|11.1 - 8.88|}{11.1} \times 100 = 19.97\%
\]

Since the percentage difference is less than 20%, the computed value is **not significantly different** from the given value.

### Conclusion:

The correct answer is:
**C. The computed value is not significantly different from the given value.**

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Would you like a breakdown of any steps or further clarification?

Here are 5 related questions:

1. How do we calculate the mean from a frequency distribution table?
2. What is the difference between sample standard deviation and population standard deviation?
3. How does increasing the number of observations in a frequency class affect the standard deviation?
4. Why do we use midpoints in grouped data to estimate the standard deviation?
5. What are the implications of a significant difference in standard deviations?

**Tip:** When handling grouped data, using midpoints provides a reasonable estimate of central tendency and spread, but it may not capture the exact nuances of individual data points.

Find the standard​ deviation, s, of the sample data summarized in the frequency distribution table below. Since individual data values are​ unavailable, instead use class midpoints to represent the data values in each class. ​Finally, compare your computed standard deviation to the standard deviation obtained from the original list of data​ values, which was 11.1.

This problem involves calculating the standard deviation from a frequency distribution table using class midpoints. Learn how to compute the sample standard deviation using essential statistical formulas, and compare it with the original data's standard deviation. Suitable for Grades 11-12 students.

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