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
Interval
30​-39
40​-49
50​-59
60​-69
70​-79
80​-89
90​-99

Frequency
2
1
4
1
12
36
37

To find the sample standard deviation \(s\) for data summarized in a frequency distribution, we use the formula:

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

Where:
- \(x\) represents the midpoint of each class interval.
- \(f\) is the frequency of each class.
- \(\bar{x}\) is the mean of the data.
- \(n\) is the total number of sample values.

### Steps:
1. **Calculate the midpoints** of each class interval.
2. **Calculate the mean (\(\bar{x}\))** of the data using the midpoints and frequencies.
3. **Calculate the variance** by finding \( f(x - \bar{x})^2 \), summing those values, and dividing by \(n-1\).
4. **Take the square root** of the variance to get the standard deviation.

Let’s break this down:

### 1. Midpoints of Class Intervals
The midpoint for each class is calculated as:

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

| Interval | Frequency \(f\) | Midpoint \(x\) |
|----------|-----------------|----------------|
| 30-39    | 2               | \( \frac{30+39}{2} = 34.5 \)  |
| 40-49    | 1               | \( \frac{40+49}{2} = 44.5 \)  |
| 50-59    | 4               | \( \frac{50+59}{2} = 54.5 \)  |
| 60-69    | 1               | \( \frac{60+69}{2} = 64.5 \)  |
| 70-79    | 12              | \( \frac{70+79}{2} = 74.5 \)  |
| 80-89    | 36              | \( \frac{80+89}{2} = 84.5 \)  |
| 90-99    | 37              | \( \frac{90+99}{2} = 94.5 \)  |

### 2. Calculate the Mean (\(\bar{x}\))
\[
\bar{x} = \frac{\sum f \cdot x}{n}
\]
Where \( n = 93 \) (the sum of all frequencies). I'll calculate this next.

Let me calculate these steps for you.The mean (\(\bar{x}\)) of the data is approximately 84.18.

### 3. Calculate the Variance
Next, we calculate the variance using the formula:

\[
\text{Variance} = \frac{\sum f(x - \bar{x})^2}{n - 1}
\]

Where \(x\) is the midpoint, and \(f\) is the frequency.

I'll calculate this now.The computed standard deviation \( s \) is approximately 12.98.

### Comparison with the original list of data
The standard deviation from the original data values was given as 11.1. The computed standard deviation from the frequency distribution (12.98) is higher, which might be due to the grouping of data in intervals leading to a loss of precision.

Would you like more details or have any questions? Here are some related questions to expand on this topic:

1. How does grouping data into intervals affect the accuracy of statistical measures?
2. What is the difference between population standard deviation and sample standard deviation?
3. How would the calculation change if the data were a population rather than a sample?
4. What are other ways to calculate the mean and standard deviation if given raw data?
5. How does skewness or kurtosis of data affect the interpretation of standard deviation?

**Tip:** When working with grouped data, the loss of precision in midpoints can sometimes cause slight discrepancies compared to calculations on the raw data.

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.

This problem demonstrates how to calculate the standard deviation from a frequency distribution table with class intervals ranging from 30-99. Learn the step-by-step process, including finding midpoints, calculating the mean, variance, and standard deviation, and comparing it to the original data values.

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