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	Frequency
30-36	2
37-43	4
44-50	4
51-57	4
58-64	14
65-71	30
72-78	39

To find the standard deviation \( s \) of the sample data summarized in the frequency distribution table, we will use the following formula:

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

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

### Step-by-Step Calculation

1. **Calculate the class midpoints** (\( x \)):
   - For the interval 30-36: \( x = \frac{30 + 36}{2} = 33 \)
   - For the interval 37-43: \( x = \frac{37 + 43}{2} = 40 \)
   - For the interval 44-50: \( x = \frac{44 + 50}{2} = 47 \)
   - For the interval 51-57: \( x = \frac{51 + 57}{2} = 54 \)
   - For the interval 58-64: \( x = \frac{58 + 64}{2} = 61 \)
   - For the interval 65-71: \( x = \frac{65 + 71}{2} = 68 \)
   - For the interval 72-78: \( x = \frac{72 + 78}{2} = 75 \)

2. **Calculate the total number of sample values** (\( n \)):
   \[
   n = 2 + 4 + 4 + 4 + 14 + 30 + 39 = 97
   \]

3. **Calculate the mean** (\( \bar{x} \)):
   \[
   \bar{x} = \frac{ \sum f \cdot x }{ n } = \frac{ (2 \cdot 33) + (4 \cdot 40) + (4 \cdot 47) + (4 \cdot 54) + (14 \cdot 61) + (30 \cdot 68) + (39 \cdot 75) }{ 97 }
   \]

   \[
   \sum f \cdot x = 66 + 160 + 188 + 216 + 854 + 2040 + 2925 = 6449
   \]

   \[
   \bar{x} = \frac{6449}{97} \approx 66.48
   \]

4. **Calculate the squared deviations** and sum them up:
   \[
   \sum f(x - \bar{x})^2
   \]
   - For the interval 30-36: \( 2(33 - 66.48)^2 \approx 2(1137.1904) = 2274.3808 \)
   - For the interval 37-43: \( 4(40 - 66.48)^2 \approx 4(705.2304) = 2820.9216 \)
   - For the interval 44-50: \( 4(47 - 66.48)^2 \approx 4(384.5904) = 1538.3616 \)
   - For the interval 51-57: \( 4(54 - 66.48)^2 \approx 4(156.2304) = 624.9216 \)
   - For the interval 58-64: \( 14(61 - 66.48)^2 \approx 14(30.1904) = 422.6656 \)
   - For the interval 65-71: \( 30(68 - 66.48)^2 \approx 30(2.3104) = 69.312 \)
   - For the interval 72-78: \( 39(75 - 66.48)^2 \approx 39(72.7104) = 2835.7056 \)

   \[
   \sum f(x - \bar{x})^2 \approx 2274.3808 + 2820.9216 + 1538.3616 + 624.9216 + 422.6656 + 69.312 + 2835.7056 \approx 10586.2688
   \]

5. **Calculate the sample standard deviation** (\( s \)):
   \[
   s = \sqrt{ \frac{ \sum f(x - \bar{x})^2 }{ n - 1 } } = \sqrt{ \frac{ 10586.2688 }{ 96 } } \approx \sqrt{110.2757} \approx 10.50
   \]

### Comparison with Given Standard Deviation

The computed standard deviation is approximately \( 10.50 \). This is close to the given standard deviation of \( 11.1 \).

Do you want more details or have any questions?

Here are some related questions you might ask:

1. How is the class midpoint calculated in a frequency distribution?
2. What is the formula for the sample mean in a frequency distribution?
3. How do you interpret the standard deviation in a frequency distribution?
4. What is the difference between sample standard deviation and population standard deviation?
5. Can you explain the concept of variance in a frequency distribution?

**Tip:** When dealing with frequency distributions, always double-check your class intervals and frequencies to ensure accuracy in your calculations.

Learn how to calculate the sample standard deviation from a frequency distribution table using class midpoints and frequencies. This problem involves statistical concepts and formulas for standard deviation. Suitable for advanced undergraduate students studying statistics.

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